Thinking Process

The general term in the expansion of $(x-a)^{n}$ i.e., $T _ {r+1}={ }^{n} C _ {r}(x)^{n-r}(-a)^{r}$. For the term independent of $x$, put $n-r=0$, then we get the value of $r$.

Solution

Given expansion is $\left(\dfrac{3 x^{2}}{2}-\dfrac{1}{3 x}\right)^{15}$

Let $T _ {r+1}$ term is the general term.

Then,

$ \begin{aligned} T _ {r+1} & ={ }^{15} C _ {r}\left({\dfrac{3 x^{2}}{2}}\right)^{15-r}\left(-\dfrac{1}{3 x}\right)^{r} \\ \\ & ={ }^{15} C _ {r} (3)^{15-r} (x)^{30-2 r} (2)^{r-15}(-1)^{r} \cdot (3)^{-r} \cdot (x)^{-r} \\ \\ & ={ }^{15} C _ {r}(-1)^{r} (3)^{15-2 r} (2)^{r-15} (x)^{30-3 r} \end{aligned} $

For independent of $x$,

$30-3 r =0 $

$3 r =30 \Rightarrow r=10 $

$\because \ T _ {r+1} =T _ {10+1}=11^{\text{th}} \text { term is independent of } x . $

$\therefore \ T _ {10+1} ={ }^{15} C _ {10}(-1)^{10} (3)^{15-20} (2)^{10-15} $

$ \quad \quad \quad \quad ={ }^{15} C _ {10} (3)^{-5} (2)^{-5} $

$ \quad \quad \quad \quad ={ }^{15} C _ {10}(6)^{-5} $

$ \quad \quad \quad \quad ={ }^{15} C _ {10} \left(\dfrac{1}{6}\right)^5 $

Solution

Given expansion is $\left(\sqrt{x}-{\dfrac{k}{x^{2}}}\right)^{10}$.

Let $T _ {r+1}$ is the general term.

Then,

$ \begin{aligned} T _ {r+1} & ={ }^{10} C _ {r}(\sqrt{x})^{10-r} \left(\dfrac{-k}{x^{2}}\right)^{r} \\ \\ & ={ }^{10} C _ {r}(x)^{[({1}/{2})(10-r)]}(-k)^{r} \cdot (x)^{-2 r} \\ \\ & ={ }^{10} C _ {r} (x)^{(5-({r}/{2}))}(-k)^{r} \cdot (x)^{-2 r} \\ \\ & ={ }^{10} C _ {r} (x)^{(5-({r}/{2})-2 r)}(-k)^{r} \\ \\ & ={ }^{10} C _ {r} (x)^{({10-5 r}/{2})}(-k)^{r} \end{aligned} $

For free from $x, \ \dfrac{10-5 r}{2}=0$

$\Rightarrow \ 10-5 r=0 \Rightarrow r=2$

Since, $T _ {2+1}=T _ 3$ is free from $x$.

$ \therefore \ T _ {2+1} ={ }^{10} C _ 2(-k)^{2}=405 $

$\Rightarrow \dfrac{10 \times 9 \times 8 !}{2 ! \times 8 !}(-k)^{2} =405 $

$\Rightarrow 45 k^{2} =405 $

$\Rightarrow k^{2}=\dfrac{405}{45}=9 $

$\therefore \ k = \pm 3$

Solution

Given, expression is $ \ (1-3 x+7 x^{2})(1-x)^{16}$.

$ \begin{aligned} & =(1-3 x+7 x^{2})[{ }^{16} C _ 0 (1)^{16}-{ }^{16} C _ 1 (1)^{15} (x)^{1}+{ }^{16} C _ 2 (1)^{14} (x)^{2}+\ldots+{ }^{16} C _ {16} (x)^{16}] \\ \\ & =(1-3 x+7 x^{2})(1-16 x+120 x^{2}+\ldots) \end{aligned} $

$\therefore \ $ Coefficient of $x=-3-16=-19$

Thinking Process

The general term in the expansion of $(x-a)^{n}$ i.e., $T _ {r+1}={ }^{n} C _ {r}(x)^{n-r}(-a)^{r}$.

Solution

Given expression is $\left(3 x-{\dfrac{2}{x^{2}}}\right)^{15}$.

Let $T _ {r+1}$ is the general term.

$ \begin{aligned} \therefore \ T _ {r+1} & ={ }^{15} C _ {r}(3 x)^{15-r} \left(\dfrac{-2}{x^{2}}\right)^{r}\\ \\ &={ }^{15} C _ {r}(3 x)^{15-r}(-2)^{r} (x)^{-2 r} \\ \\ & ={ }^{15} C _ {r} (3)^{15-r} (x)^{15-3 r}(-2)^{r} \end{aligned} $

For independent of $x, \ 15-3 r=0 \Rightarrow r=5$

Since, $T _ {5+1}=T _ 6$ is independent of $x$.

$ \begin{aligned} \therefore \quad T _ {5+1} & ={ }^{15} C _ 5 (3)^{15-5}(-2)^{5} \\ \\ & =-\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !} \cdot 3^{10} \cdot 2^{5} \\ \\ & =-3003 \cdot 3^{10} \cdot 2^{5} \end{aligned} $

Thinking Process

In the expansion of $(a+b)^{n}$, if $n$ is even, then this expansion has only one middle term i.e., $\left(\dfrac{n}{2}+1\right)^{\text{th}}$ term is the middle term and if $n$ is odd, then this expansion has two middle terms i.e., $\left(\dfrac{n+1}{2}\right)^{\text{th}}$ and $\left(\dfrac{n+1}{2}+1\right)^{\text{th}}$ are two middle terms.

Solution

(i) Given expression is $\left(\dfrac{x}{a}-\dfrac{a}{x}\right)^{10}$

Here, the power of Binomial i.e., $n=10$ is even.

Since, it has one middle term $\left(\dfrac{10}{2}+1\right)^{\text{th}}$ term i.e., $6^{\text{th}}$ term.

$ \begin{aligned} \therefore \quad T _ 6 & =T _ {5+1}={ }^{10} C _ 5 \left(\dfrac{x}a \right)^{10-5} \left(\dfrac{-a}{x}\right)^5 \\ \\ & =-{ }^{10} C _ 5 \left(\dfrac{x}a \right)^{5} \left(\dfrac{a}x\right)^{5} \\ \\ & =-\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 5 \times 4 \times 3 \times 2 \times 1} \ \left(\dfrac{x}{a}\right)^5 \ \left(\dfrac{x}{a}\right)^{-5} \\ \\ & =-9 \times 4 \times 7=-252 \end{aligned} $

(ii) Given expression is $\left(3 x-{\dfrac{x^{3}}{6}}\right)^{9}$.

Here, $n=9 \quad$ [odd]

Since, the Binomial expansion has two middle terms i.e., $\left(\dfrac{9+1}{2}\right)^{\text{th}}$ and $\left(\dfrac{9+1}{2}+1\right)^{\text{th}}$ i.e., $5^{\text{th}}$ term and $6^{\text{th}}$ term.

$ \begin{aligned} \therefore \quad T _ 5 & =T _ {(4+1)}={ }^{9} C _ 4(3 x)^{9-4}\left(-\dfrac{x^{3}}{6}\right)^4 \\ \\ & =\dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !} 3^{5} x^{5} x^{12} 6^{-4} \\ \\ & =\dfrac{7 \times 6 \times 3 \times 3^{1}}{2^{4}} x^{17}=\dfrac{189}{8} x^{17} \end{aligned} $

$ \begin{aligned} \therefore \quad T _ 6 & =T _ {5+1}={ }^{9} C _ 5(3 x)^{9-5}\left(-\dfrac{x^{3}}{6}\right)^5 \\ \\ & =-\dfrac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1} \cdot 3^{4} \cdot x^{4} \cdot x^{15} \cdot 6^{-5} \\ \\ & =\dfrac{-21 \times 6}{3 \times 2^{5}} x^{19}=\dfrac{-21}{16} x^{19} \end{aligned} $

Solution

Given expression is $(x-x^{2})^{10}$.

Let the term $T _ {r+1}$ is the general term.

$ \begin{aligned} \therefore \quad T _ {r+1} & ={ }^{10} C _ {r} \cdot (x)^{10-r} \cdot (-x^{2})^{r} \\ \\ & =(-1)^{r} \cdot{ }^{10} C _ {r} \cdot (x)^{10-r} \cdot (x)^{2 r} \\ \\ & =(-1)^{r} \cdot { }^{10} C _ {r} \cdot (x)^{10+r} \end{aligned} $

For the coefficient of $x^{15}$,

$10+r =15 \Rightarrow r=5 $

$T _ {5+1} =(-1)^{5} \cdot { }^{10} C _ 5 \cdot (x)^{15} $

$\therefore \ \text { Coefficient of } x^{15} =(-1)^5 \cdot {}^{10} C _ 5 =(-1) \times \dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 !}{5 \times 4 \times 3 \times 2 \times 1 \times 5 !} $

$ =-3 \times 2 \times 7 \times 6=-252$

Thinking Process

In this type of questions, first of all find the general terms, in the expansion $(x-y)^{n}$ using the formula $T _ {r+1}={ }^{n} C _ {r} x^{n-r}(-y)^{r}$ and then put $n-r$ equal to the required power of $x$ of which coefficient is to be find out.

Solution

Given expression is $\left(x^{4}-{\dfrac{1}{x^{3}}}\right)^{15}$.

Let the term $T _ {r+1}$ contains the coefficient of $\dfrac{1}{x^{17}}$ i.e., $x^{-17}$.

$\therefore \ T _ {r+1} ={ }^{15} C _ {r}(x^{4})^{15-r} \left(-\dfrac{1}{x^{3}}{ }\right)^{r}$

$ ={ }^{15} C _ {r} (x)^{60-4 r}(-1)^{r} (x)^{-3 r} $

$={ }^{15} C _ {r} (x)^{60-7 r}(-1)^{r}$

For the coefficient $x^{-17}$,

$ 60-7 r=-17 $

$ 7 r=77 \Rightarrow r=11 $

Now, $ \ T _ {11+1}={ }^{15} C _ {11} \cdot (x)^{60-77} \cdot (-1)^{11} $

$\therefore \ \text { Coefficient of } x^{-17} =(-1)^{11} \cdot {}^{15} C _ {11} =\dfrac{-15 \times 14 \times 13 \times 12 \times 11 !}{11 ! \times 4 \times 3 \times 2 \times 1} $

$ =-15 \times 7 \times 13=-1365$

Solution

Given expression is $(y^{({1}/{2})}+x^{({1}/{3})})^{n}$.

The sixth term of this expansion is

$ T _ 6=T _ {5+1}={ }^{n} C _ 5 \cdot (y^{({1}/{2})})^{n-5} \cdot (x^{({1}/{3})})^{5} \quad \ldots (i) $

Now, given that the Binomial coefficient of the third term from the end is 45 .

We know that, Binomial coefficient of third term from the end $=$ Binomial coefficient of third term from the begining $={ }^{n} C _ 2$

$ \begin{aligned} & \because \ { }^{n} C _ 2=45 \\ \\ & \Rightarrow \ \dfrac{n(n-1)(n-2) !}{2 !(n-2) !}=45 \\ \\ & \Rightarrow \ n(n-1)=90 \\ \\ & \Rightarrow \ n^{2}-n-90=0 \\ \\ & \Rightarrow \ n^{2}-10 n+9 n-90=0 \\ \\ & \Rightarrow \ n(n-10)+9(n-10)=0 \\ \\ & \Rightarrow \ (n-10)(n+9)=0 \\ \\ & \Rightarrow \ (n+9)=0 \text { or }(n-10)=0 \\ \\ & \therefore \ n=10 \quad[\because n \neq-9] \end{aligned} $

From Eq. (i),

$ T _ 6={ }^{10} C _ 5 \cdot (y)^{5 / 2} \cdot (x)^{5 / 3}=252 (y)^{5 / 2} \cdot (x)^{5 / 3} $

Thinking Process

Coefficient of $(r+1)^{\text{th}}$ term in the expansion of $(1+x)^{n}$ is ${ }^{n} C _ {r}$. Use this formula to solve the above problem.

Solution

Given expression is $(1+x)^{18}$.

Now, $(2 r+4)^{\text{th}}$ term i.e., $T _ {2 r+3+1}$.

$ \begin{aligned} \therefore \quad T _ {2 r+3+1} & ={ }^{18} C _ {2 r+3} \cdot (1)^{18-2 r-3} \cdot (x)^{2 r+3} \\ \\ & ={ }^{18} C _ {2 r+3} \cdot (x)^{2 r+3} \end{aligned} $

Now, $ \ (r-2)^{\text{th}}$ term i.e., $T _ {r-3+1}$.

$ \therefore \ T _ {r-3+1} ={}^{18} C _ {r-3} \cdot (x)^{r-3} $

As, $ \ {}^{18} C _ {2 r+3} = {}^{18} C _ {r-3} $

$\Rightarrow 2 r+3+r-3 =18 $ $\qquad [\because \ {}^nC _ x = {}^nC _ y \Rightarrow x+y=n]$

$\Rightarrow 3 r =18 $

$ \therefore \ r =6 $

Thinking Process

In the expansion of $(x+y)^{n}$, the coefficient of $(r+1)^{\text{th}}$ term is ${ }^{n} C _ {r}$. Use this formula to get the required coefficient.

If $a, b$ and $c$ are in AP, then $2 b=a+c$.

Solution

Given expression is $(1+x)^{2 n}$.

Now, coefficient of $2^{\text{nd}}$ term $={ }^{2 n} C _ 1$

Coefficient of $3^{\text{rd}}$ term $={ }^{2 n} C _ 2$

Coefficient of $4^{\text{th}}$ term $={ }^{2 n} C _ 3$

Given that, ${ }^{2 n} C _ 1,{ }^{2 n} C _ 2$ and ${ }^{2 n} C _ 3$ are in AP.

Then, $ \ 2 \cdot{ }^{2 n} C _ 2={ }^{2 n} C _ 1+{ }^{2 n} C _ 3$

$\Rightarrow \ 2 \Big[\dfrac{2 n(2 n-1)(2 n-2) !}{2 \times 1 \times(2 n-2) !}\Big]=\dfrac{2 n(2 n-1) !}{(2 n-1) !}+\dfrac{2 n(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !}$

$\Rightarrow \ n(2 n-1)=n+\dfrac{n(2 n-1)(2 n-2)}{6}$

$\Rightarrow \ n(12 n-6)=n(6+4 n^{2}-4 n-2 n+2)$

$\Rightarrow \ 12 n-6=(4 n^{2}-6 n+8)$

$\Rightarrow \ 6(2 n-1)=2(2 n^{2}-3 n+4)$

$\Rightarrow \ 3(2 n-1)=2 n^{2}-3 n+4$

$\Rightarrow \ 2 n^{2}-3 n+4-6 n+3=0$

$\Rightarrow \ 2 n^{2}-9 n+7=0$

Hence proved.

Solution

Given, expression $=(1+x+x^{2}+x^{3})^{11}=[(1+x)+x^{2}(1+x)]^{11}$

$ =[(1+x)(1+x^{2})]^{11}=(1+x)^{11} \cdot(1+x^{2})^{11} $

Now, above expansion becomes

$ \begin{aligned} & =({ }^{11} C _ 0+{ }^{11} C _ 1 x+{ }^{11} C _ 2 x^{2}+{ }^{11} C _ 3 x^{3}+{ }^{11} C _ 4 x^{4}+\ldots)({ }^{11} C _ 0+{ }^{11} C _ 1 x^{2}+{ }^{11} C _ 2 x^{4}+\ldots) \\ \\ & =(1+11 x+55 x^{2}+165 x^{3}+330 x^{4}+\ldots)(1+11 x^{2}+55 x^{4}+\ldots) \end{aligned} $

$\therefore$ Coefficient of $x^{4}=55+605+330=990$

Solution

Given expression is $\left(\dfrac{p}{2}+2\right)^{8}$.

Here, $n=8 \quad$ [even]

Since, this Binomial expansion has only one middle term i.e., $\left(\dfrac{8}{2}+1\right)^{\text{th}}$ $=5^{\text{th}}$ term

$ T _ 5 =T _ {4+1}={ }^{8} C _ 4 \cdot \left(\dfrac{p}2\right)^{8-4} \cdot 2^{4} $

$ \Rightarrow 1120 ={ }^{8} C _ 4 \cdot (p)^{4} \cdot (2)^{-4} \cdot (2)^{4} $

$ \Rightarrow 1120 =\dfrac{8 \times 7 \times 6 \times 5 \times 4 !}{4 ! \times 4 \times 3 \times 2 \times 1} \times p^{4} $

$\Rightarrow 1120=7 \times 2 \times 5 \times p^{4} $

$\Rightarrow p^{4}=\dfrac{1120}{70}=16 \Rightarrow p^{4}=2^{4}$

$\Rightarrow p^{2}=4 \Rightarrow p= \pm 2$

Solution

Given, expansion is $\left(x-\dfrac{1}x \right)^{2 n}$.

This Binomial expansion has even power. So, this has one middle term.

i.e., $ \ \left(\dfrac{2 n}{2}+1\right)^{\text{th}} \text { term }=(n+1)^{\text{th}} \text { term }$

$T _ {n+1} ={ }^{2 n} C _ {n} \cdot (x)^{2 n-n} \cdot \left(-\dfrac{1}{x}{ }\right)^{n}={ }^{2 n} C _ {n} \cdot (x)^{n} \cdot (-1)^{n} \cdot (x)^{-n}$

$={ }^{2 n} C _ {n} \cdot (-1)^{n}=(-1)^{n} \cdot \dfrac{(2 n) !}{n ! n !}=\dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(2 n-1)(2 n)}{n ! n !} \cdot(-1)^{n}$

$=\dfrac{1 \cdot 3 \cdot 5 \ldots(2 n-1) \cdot 2 \cdot 4 \cdot 6 \ldots(2 n)}{1 \cdot 2 \cdot 3 \cdot \ldots n(n !)} \cdot (-1)^{n} $

$=\dfrac{1 \cdot 3 \cdot 5 \ldots(2 n-1) \cdot 2^{n}(1 \cdot 2 \cdot 3 \ldots n)(-1)^{n}}{(1 \cdot 2 \cdot 3 \ldots n)(n !)} $

$=\dfrac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)]}{n !} \cdot (-2)^{n}$

Hence proved.

Solution

Here, the Binomial expression is $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}{ }\right)^{n}$.

Now, $7^{\text{th}}$ term from beginning $T _ 7=T _ {6+1}={ }^{n} C _ 6 \cdot (\sqrt[3]{2})^{n-6} \left(\dfrac{1}{\sqrt[3]{3}}\right)^{6} \quad \ldots (i)$

and $7^{\text{th}}$ term from end i.e., $T _ 7$ from the beginning of $\left(\dfrac{1}{\sqrt[3]{3}}+\sqrt[3]{2}{ }\right)^{n}$

i.e., $ \ T _ 7={ }^{n} C _ 6 \cdot \left(\dfrac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^{6}$

Given that, $ \ \dfrac{{ }^{n} C _ 6 \cdot (\sqrt[3]{2})^{n-6} \left(\dfrac{1}{\sqrt[3]{3}}\right)^6}{{ }^{n} C _ 6 \cdot \left(\dfrac{1}{\sqrt[3]{3}}\right)^{n-6}(\sqrt[3]{2})^{6}}=\dfrac{1}{6} $

$\Rightarrow \dfrac{2^{({{(n-6)}/{3}})} \cdot 3^{(-6 / 3)}}{3^{(-({n-6})/{3})} \cdot 2^{(6 / 3)}}=\dfrac{1}{6}$

$\Rightarrow \quad \bigg[ 2^{\left({(n-6)}/{3}\right)} \cdot 2^{\left({-6}/{3}\right)}\bigg] \bigg[3^{\left({-6}/{3}\right)} \cdot 3^{\left({(n-6)}/{3}\right)}\bigg]=6^{-1}$

$ \begin{aligned} & \Rightarrow \quad 2^{({(n-6)}/{3}-\dfrac{6}{3})} \cdot 3^{({(n-6)}/{3}-\dfrac{6}{3})}=6^{-1} \\ \\ & \Rightarrow \quad (2 \cdot 3)^{(({n}/{3})-4)}=(2 \cdot 3)^{-1} \\ \\ & \Rightarrow \quad \dfrac{n}{3}-4=-1 \Rightarrow \dfrac{n}{3}=3 \\ \\ & \therefore \quad n=9 \end{aligned} $

Solution

(i) Given expression is $(x+a)^{n}$.

$\therefore \ (x+a)^{n}={ }^{n} C _ 0 x^{n} a^{0}+{ }^{n} C _ 1 x^{n-1} a^{1}+{ }^{n} C _ 2 x^{n-2} a^{2}+{ }^{n} C _ 3 x^{n-3} a^{3}+\ldots+{ }^{n} C _ {n} a^{n}$

Now, sum of odd terms

i.e.,

$ \ O={ }^{n} C _ 0 x^{n} a^0+{ }^{n} C _ 2 x^{n-2} a^{2}+{}^nC _ 4 x^{n-4} a^4 +\ldots $

and sum of even terms

i.e.,

$ \ E={ }^{n} C _ 1 x^{n-1} a^1+{ }^{n} C _ 3 x^{n-3} a^{3}+{}^nC _ 5 x^{n-5} a^5+\ldots $

$\because \ (x+a)^{n}=O+E \quad \ldots (i)$

Similarly, $\ (x-a)^{n}=O-E \quad \ldots (ii)$

$\therefore \ (O+E)(O-E)=(x+a)^{n}(x-a)^{n}$ $\quad $ [on multiplying Eqs. (i) and (ii)]

$\Rightarrow \ O^{2}-E^{2}=(x^{2}-a^{2})^{n}$

Hence proved.

(ii) $4 O E=(O+E)^{2}-(O-E)^{2}=[(x+a)^{n}]^{2}-[(x-a)^{n}]^{2}$

[from Eqs. (i) and (ii)]

$ 4 OE=(x+a)^{2 n}-(x-a)^{2 n} $

Hence proved.

Solution

Given expression is $\left(x^{2}+\dfrac{1}x\right)^{2 n}$.

Let $x^{p}$ occur in the expansion of $\left(x^{2}+\dfrac{1}x\right)^{2 n}$.

$T _ {r+1} ={ }^{2 n} C _ {r}(x^{2})^{2 n-r} \left(\dfrac{1}{x}\right)^r $

$={ }^{2 n} C _ {r} (x)^{4 n-2 r} (x)^{-r}={ }^{2 n} C _ {r} (x)^{4 n-3 r} $

Let $\ 4 n-3 r =p$

$\Rightarrow \ 3 r =4 n-p $

$\Rightarrow \ r=\dfrac{4 n-p}{3} $

$\therefore \ \text { Coefficient of } x^{p} ={ }^{2 n} C _ {r}=\dfrac{(2 n) !}{r !(2 n-r) !}=\dfrac{(2 n) !}{\left(\dfrac{4 n-p}{3}\right) ! \left(2 n-\dfrac{4 n-p}{3}\right) !} $

$ =\dfrac{(2 n) !}{\left(\dfrac{4 n-p}{3}\right) ! \left(\dfrac{6 n-4 n+p}{3} \right)!}=\dfrac{(2 n) !}{\left(\dfrac{4 n-p}{3}\right) ! \left(\dfrac{2 n+p}{3}\right) !}$

Solution

Given expression is $(1+x+2 x^{3}) \left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$.

Now, consider $\left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$

$ \begin{aligned} T _ {r+1} & ={ }^{9} C _ {r} \left(\dfrac{3x^{2}}{2}\right)^{9-r} \left(-\dfrac{1}{3 x}\right)^r \\ \\ & ={ }^{9} C _ {r} \left(\dfrac{3}2\right)^{9-r} (x)^{18-2 r}\left(-\dfrac{1}3\right)^{r} (x)^{-r}={ }^{9} C _ {r} \left(\dfrac{3}2\right)^{9-r}\left(-\dfrac{1}3\right)^{r} (x)^{18-3 r} \end{aligned} $

Hence, the general term in the expansion of $(1+x+2 x^{3}) \left(\dfrac{3}{2} x^{2}-\dfrac{1}{3 x}\right)^{9}$

$ ={ }^{9} C _ {r} \left(\dfrac{3}2\right)^{9-r}\left(-\dfrac{1}3\right)^{r} (x)^{18-3 r}+{ }^{9} C _ {r} \left(\dfrac{3}2\right)^{9-r}\left(-\dfrac{1}3\right)^{r} (x)^{19-3 r}+2 \cdot{ }^{9} C _ {r} \left(\dfrac{3}2\right)^{9-r}\left(-\dfrac{1}3\right)^{r} (x)^{21-3 r} $

For term independent of $x$, putting $18-3 r=0,19-3 r=0$ and $21-3 r=0$, we get

$ r=6, r=\dfrac{19}{3}, r=7 $

Since, the possible value of $r$ are 6 and 7 .

Hence, second term is not independent of $x$.

$\therefore \ $ The term independent of $x= T _ 7+T _ 8$

$ \ \ ={ }^{9} C _ 6 \left(\dfrac{3}2\right)^{9-6}\left(-\dfrac{1}3\right)^{6}+2 \cdot{ }^{9} C _ 7 \left(\dfrac{3}2\right)^{9-7}\left(-\dfrac{1}3\right)^{7}$

$ \begin{aligned} & =\dfrac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 \times 2} \cdot \dfrac{3^{3}}{2^{3}} \cdot \dfrac{1}{3^{6}}-2 \cdot \dfrac{9 \times 8 \times 7 !}{7 ! \times 2 \times 1} \cdot \dfrac{3^{2}}{2^{2}} \cdot \dfrac{1}{3^{7}} \\ \\ & =\dfrac{84}{8} \cdot \dfrac{1}{3^{3}}-\dfrac{36}{4} \cdot \dfrac{2}{3^{5}} \\ \\ &=\dfrac{7}{18}-\dfrac{2}{27}=\dfrac{21-4}{54}=\dfrac{17}{54} \end{aligned} $

Solution

Option (c) Here, $(x+a)^{100}+(x-a)^{100}$

Total number of terms is $102$ in the expansion of $(x+a)^{100}+(x-a)^{100}$

$50$ terms of $(x+a)^{100}$ cancel out $50$ terms of $(x-a)^{100} .51$ terms of $(x+a)^{100}$ get added to the $51$ terms of $(x-a)^{100}$.

Alternate Method

$ \begin{aligned} (x+a)^{100}+(x-a)^{100} & ={ }^{100} C _ 0 x^{100}+{ }^{100} C _ 1 x^{99} a+\ldots+{ }^{100} C _ {100} a^{100} \\ \\ & +{ }^{100} C _ 0 x^{100}-{ }^{100} C _ 1 x^{99} a+\ldots+{ }^{100} C _ {100} a^{100} \\ \\ & =2 \underbrace{.[{ }^{100} C _ 0 x^{100}+{ }^{100} C _ 2 x^{98} a^{2}+\ldots+{ }^{100} C _ {100} a^{100}]} _ {51 \text { terms }} \end{aligned} $

• Option (a) 50: This option is incorrect because the total number of terms in the expansion of $(x+a)^{100} + (x-a)^{100}$ after simplification is not $50$. The correct number of terms is $51$, as the odd-powered terms cancel out and only the even-powered terms remain.

• Option (b) 202: This option is incorrect because it suggests that the total number of terms in the expansion is $202$, which is not possible. The original expansions of $(x+a)^{100}$ and $(x-a)^{100}$ each have $101$ terms, but after simplification, the number of terms is reduced to $51$ due to the cancellation of odd-powered terms.

• Option (d) None of these: This option is incorrect because there is a correct answer provided in the options, which is $51$. Therefore, “None of these” is not the correct choice.

Thinking Process

In the expansion of $(x+y)^{n}$, the coefficient of $(r+1)$ th term is ${ }^{n} C _ {r}$.

Solution

Option (a) Given that, $r>1, n>2$ and the coefficients of $(3 r)^{th}$ and $(r+2)^{th}$ term are equal in the expansion of $(1+x)^{2 n}$.

$ \begin{aligned} & \text { Then, } \quad T _ {3 r}=T _ {3 r-1+1}={ }^{2 n} C _ {3 r-1} x^{3 r-1} \\ \\ & \text { and } \quad T _ {r+2}=T _ {r+1+1}={ }^{2 n} C _ {r+1} x^{r+1} \\ \\ & \text { Given, } \quad{ }^{2 n} C _ {3 r-1}={ }^{2 n} C _ {r+1} \quad\left[\because{ }^n C _ x={ }^n C _ y \Rightarrow x+y=n\right] \\ \\ & \Rightarrow \quad 3 r-1+r+1=2 n \\ \\ & \Rightarrow \quad 4 r=2 n \Rightarrow n=\dfrac{4 r}{2} \\ \\ & \therefore \quad n=2 r \\ \\
\end{aligned} $

• Option (b) $n=3r$: is incorrect because, according to the given condition, the equation derived from the equality of the binomial coefficients is $4r = 2n$, which simplifies to $n = 2r$. This does not match $n = 3r$.

• Option (c) $n=2r+1$: is incorrect because, as derived from the equality of the binomial coefficients, the correct relationship is $4r = 2n$, which simplifies to $n = 2r$. This does not match $n = 2r + 1$.

• Option (d) None of these: is incorrect because the correct answer, as derived from the given conditions and the binomial coefficient equality, is $n = 2r$. Therefore, there is a correct option provided, making “None of these” incorrect.

Solution

Option (c) Let two successive terms in the expansion of $(1+x)^{24}$ are $(r+1) ^{th} $ and $(r+2)^{th}$ terms.

$ \therefore $ $ \ T _ {r+1} ={ }^{24} C _ {r} x^{r} $

and $T _ {r+2} ={ }^{24} C _ {r+1} x^{r+1} $

$\dfrac{{ }^{24} C _ {r}}{{ }^{24} C _ {r+1}} =\dfrac{1}{4} $

Given that, $ \ \dfrac{\frac{(24) !}{r !(24-r) !}}{(24) !} =\dfrac{1}{4} $

$ \Rightarrow \dfrac{(r+1) !(24-r-1) !}{(r+1) r(23-r) !} =\dfrac{1}{4} $

$\Rightarrow \dfrac{(r+124-r)(23-r) !}{r 4-r} =\dfrac{1}{4} $

$\Rightarrow 4 r+4=24-r $

$\Rightarrow 5 r=20 \Rightarrow r=4 $

$ \therefore T _ {4+1} =T _ 5 \text { and } T _ {4+2}=T _ 6 $

Hence, 5th and 6th terms.

• Option (a) 3rd and 4th: The coefficients of the 3rd and 4th terms in the expansion of $(1+x)^{24}$ are not in the ratio 1:4. When we calculate the ratio of the coefficients of these terms, it does not satisfy the given condition.

• Option (b) 4th and 5th: The coefficients of the 4th and 5th terms in the expansion of $(1+x)^{24}$ are not in the ratio 1:4. When we calculate the ratio of the coefficients of these terms, it does not satisfy the given condition.

• Option (d) 6th and 7th: The coefficients of the 6th and 7th terms in the expansion of $(1+x)^{24}$ are not in the ratio 1:4. When we calculate the ratio of the coefficients of these terms, it does not satisfy the given condition.

Solution

Option (d) $\because \ $ Coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}={ }^{2 n} C _ {n}$ and coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n-1}={ }^{2 n-1} C _ {n}$

$ \dfrac{{ }^{2 n} C _ {n}}{{ }^{2 n-1} C _ {n}} =\dfrac{\dfrac{(2 n) !}{n ! n !}}{\dfrac{(2 n-1) !}{n !(n-1) !}} $

$ =\dfrac{(2 n) ! n !(n-1) !}{n ! n !(2 n-1) !} $

$ =\dfrac{2 n(2 n-1) ! n !(n-1) !}{n ! n(n-1) !(2 n-1) !} $

$ =\dfrac{2 n}{n}=\dfrac{2}{1}=2: 1 $

• Option (a) $1: 2$: is incorrect because the ratio of the coefficients is $\dfrac{2n}{n} = 2:1$, not $1:2$.

• Option (b) $1: 3$: is incorrect because the ratio of the coefficients is $\dfrac{2n}{n} = 2:1$, not $1:3$.

• Option (c) $3: 1$: is incorrect because the ratio of the coefficients is $\dfrac{2n}{n} = 2:1$, not $3:1$.

Solution

Option (b) The expansion of $(1+x)^{n}$ is ${ }^{n} C _ 0+{ }^{n} C _ 1 x+{ }^{n} C _ 2 x^{2}+{ }^{n} C _ 3 x^{3}+\ldots+{ }^{n} C _ {n} x^{n}$

$\therefore \quad$ Coefficient of $2^{nd}$ term $={ }^{n} C _ 1$,

Coefficient of $3^{rd}$ term $={ }^{n} C _ 2$

and coefficient of $4^{th}$ term $={ }^{n} C _ 3$

Given that, ${ }^{n} C _ 1,{ }^{n} C _ 2$ and ${ }^{n} C _ 3$ are in AP

$\therefore \ 2{ }^{n} C _ 2={ }^{n} C _ 1+{ }^{n} C _ 3 $

$\Rightarrow \ 2 \dfrac{(n) !}{(n-2) ! 2 !}=\dfrac{(n) !}{(n-1) !}+\dfrac{(n) !}{3 !(n-3) !} $

$\Rightarrow \ \dfrac{2 \cdot n(n-1)(n-2) !}{(n-2) ! 2 !}=\dfrac{n(n-1) !}{(n-1) !}+\dfrac{n(n-1)(n-2)(n-3) !}{3 \cdot 2 \cdot 1(n-3) !} $

$\Rightarrow \ n(n-1)=n+\dfrac{n(n-1)(n-2)}{6} $

$\Rightarrow \ 6 n-6=6+n^{2}-3 n+2 $

$\Rightarrow \ n^{2}-9 n+14=0 $

$\Rightarrow \ n(n-7)-2(n-7)=0 $

$\Rightarrow \ (n-7)(n-2)=0 $

$\Rightarrow \ n=2 \text { or } n=7$

Since, $n=2$ is not possible.

$ \therefore \ n=7 $

• Option (a) 2: This option is incorrect because when $ n = 2 $, the coefficients of the 2nd, 3rd, and 4th terms in the expansion of $(1+x)^n$ do not form an arithmetic progression (AP). Specifically, the expansion $(1+x)^2$ yields the terms $1 + 2x + x^2$, and there is no 4th term. Therefore, the condition of the coefficients being in AP cannot be satisfied.

• Option (c) 11: This option is incorrect because when $ n = 11 $, the coefficients of the 2nd, 3rd, and 4th terms in the expansion of $(1+x)^{11}$ do not form an arithmetic progression (AP). Specifically, the coefficients are ${}^{11}C _ 1 = 11$, ${}^{11}C _ 2 = 55$, and ${}^{11}C _ 3 = 165$. These values do not satisfy the condition $2 \cdot {}^{11}C _ 2 = {}^{11}C _ 1 + {}^{11}C _ 3$.

• Option (d) 14: This option is incorrect because when $ n = 14 $, the coefficients of the 2nd, 3rd, and 4th terms in the expansion of $(1+x)^{14}$ do not form an arithmetic progression (AP). Specifically, the coefficients are ${}^{14}C _ 1 = 14$, ${}^{14}C _ 2 = 91$, and ${}^{14}C _ 3 = 364$. These values do not satisfy the condition $2 \cdot {}^{14}C _ 2 = {}^{14}C _ 1 + {}^{14}C _ 3$.

Solution

Option (b) Since, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n}$ is ${ }^{2 n} C _ {n}$.

$\therefore \quad A={ }^{2 n} C _ {n}$

Now, the coefficient of $x^{n}$ in the expansion of $(1+x)^{2 n-1}$ is ${ }^{2 n-1} C _ {n}$.

$\therefore \quad B={ }^{2 n-1} C _ {n}$

Now,

$ \dfrac{A}{B}=\dfrac{{ }^{2 n} C _ {n}}{{ }^{2 n-1} C _ {n}}=\dfrac{2}{1}=2 $

• Option (a) 1: This option is incorrect because the ratio of the binomial coefficients $\dfrac{{}^{2n}C _ n}{{}^{2n-1}C _ n}$ is not equal to 1. The correct ratio is 2, as shown in the solution.

• Option (c) $\dfrac{1}{2}$: This option is incorrect because the ratio of the binomial coefficients $\dfrac{{}^{2n}C _ n}{{}^{2n-1}C _ n}$ is not equal to $\dfrac{1}{2}$. The correct ratio is 2, as shown in the solution.

• Option (d) $\dfrac{1}{n}$: This option is incorrect because the ratio of the binomial coefficients $\dfrac{{}^{2n}C _ n}{{}^{2n-1}C _ n}$ is not dependent on $n$ in such a way that it would result in $\dfrac{1}{n}$. The correct ratio is 2, as shown in the solution.

Solution

Option (c) Given expansion is $\left(\dfrac{1}{x}+x \sin x\right)^{10}$.

Since, $n=10$ is even, so this expansion has only one middle term i.e., $6^{th}$ term.

$ \begin{aligned} & \therefore \quad T _ 6=T _ {5+1}={ }^{10} C _ 5 (\dfrac{1}{x}{ })^{10-5}(x \sin x)^{5} \\ \\ & \Rightarrow \quad \dfrac{63}{8}={ }^{10} C _ 5 x^{-5} x^{5} \sin ^{5} x \\ \\ & \Rightarrow \quad \dfrac{63}{8}=\dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 !}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 5 !} \sin ^{5} x \\ \\ & \Rightarrow \quad \dfrac{63}{8}=2 \cdot 9 \cdot 2 \cdot 7 \cdot \sin ^{5} x \\ \\ & \Rightarrow \quad \sin ^{5} x=\dfrac{1}{32} \\ \\ & \Rightarrow \quad \sin ^{5} x=\left(\dfrac{1}2\right)^{5} \\ \\ & \Rightarrow \quad \sin x=\dfrac{1}{2} \\ \\ & \therefore \quad x=n \pi+(-1)^{n} \dfrac{\pi}{6} \end{aligned} $

• Option (a) $2 n \pi+\dfrac{\pi}{6}$ : is incorrect because it does not account for the alternating sign that arises from the periodicity of the sine function. The sine function $\sin x = \dfrac{1}{2}$ occurs at $x = n \pi + (-1)^n \dfrac{\pi}{6}$, not at $2 n \pi + \dfrac{\pi}{6}$.

• Option (b) $n \pi+\dfrac{\pi}{6}$ : is incorrect because it only considers the positive solution for $\sin x = \dfrac{1}{2}$ and does not account for the alternating sign that occurs due to the periodic nature of the sine function. The correct general solution should include the $(-1)^n$ factor to account for both positive and negative solutions.

• Option (d) $n \pi+(-1)^{n} \dfrac{\pi}{3}$ : is incorrect because $\sin x = \dfrac{1}{2}$ does not occur at $x = n \pi + (-1)^n \dfrac{\pi}{3}$. The correct angles where $\sin x = \dfrac{1}{2}$ are $x = n \pi + (-1)^n \dfrac{\pi}{6}$.

Thinking Process

In the expansion of $(1+x)^{n}$, the largest coefficient is ${ }^{n} C _ {n / 2}$ (when $n$ is even).

Solution

Largest coefficient in the expansion of $(1+x)^{30}={ }^{30} C _ {30 / 2}={ }^{30} C _ {15}$

Solution

Given expansion is $(x+y+z)^{n}=[x+(y+z)]^{n}$.

$[x+(y+z)]^{n}={ }^{n} C _ 0 x^{n}+{ }^{n} C _ 1 x^{n-1}(y+z)+{ }^{n} C _ 2 x^{n-2}(y+z)^{2}+\ldots+{ }^{n} C _ {n}(y+z)^{n}$

$\therefore \quad$ Number of terms $=1+2+3+\ldots+n+(n+1)$

$ =\dfrac{(n+1)(n+2)}{2} $

Solution

Let constant be $T _ {r+1}$.

$\therefore T _ {r+1} ={ }^{16} C _ {r}(x^{2})^{16-r}(-\dfrac{1}{x^{2}})^r $

$ ={ }^{16} C _ {r} x^{32-2 r}(-1)^{r} x^{-2 r} $

$ ={ }^{16} C _ {r} x^{32-4 r}(-1)^{r} $

$ \text { For constant term, } \quad 32-4 r =0 \Rightarrow r=8 $

$\therefore T _ {8+1} ={ }^{16} C _ 8 $

Solution

Given expansions is $\left(\sqrt[3]{2}+\dfrac{1}{\sqrt[3]{3}}\right)^{n}$

$ \therefore \quad \quad T _ 7=T _ {6+1}={ }^{n} C _ 6(\sqrt[3]{2})^{n-6} \left(\dfrac{1}{\sqrt[3]{3}}\right)^{6} $

Since, $T _ 7$ from end is same as the $T _ 7$ from beginning of $\left(\dfrac{1}{\sqrt[3]{3}}+\sqrt[3]{2}{ }\right)^{n}$

Then,

$ T _ 7={ }^{n} C _ 6 \left(\dfrac{1}{\sqrt[3]{3}}\right)^{n-6}\left(\sqrt[3]{2}\right)^{6} $

Given that,

$ { }^{n} C _ 6(2)^{{(n-6)}/{3}}(3)^{-6 / 3}={ }^{n} C _ 6(3)^{(-{(n-6)}/{3})} 2^{({6}/{3}) } $

$ \Rightarrow \quad \text { 2 }^ {((n-12)/{3})}={\left(\dfrac{1}{\dfrac{1}{3}}\right)}^{(({n-12})/{3})} $

which is true, when $\dfrac{n-12}{3}=0$.

$ \Rightarrow n-12=0 \Rightarrow n=12 $

Thinking Process

In the expansion of $(x-a)^{n}, T _ {r+1}={ }^{n} C _ {r} x^{n-r}(-a)^{r}$

Solution

Given expansion is $\left(\dfrac{1}{a}-\dfrac{2 b}3\right)^{10}$.

Let $T _ {r+1}$ has the coefficient of $a^{-6} b^{4}$.

$\therefore \quad T _ {r+1}={ }^{10} C _ {r} \left(\dfrac{1}a\right)^{10-r}-\left(\dfrac{2 b}3\right)^{r}$

For coefficient of $a^{-6} b^{4}, \quad 10-r=6 \Rightarrow r=4$

Coefficient of $a^{-6} b^{4}={ }^{10} C _ 4\left(\dfrac{-2}{3}\right)^{4}$

$\therefore \quad=\dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 !}{6 ! \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \dfrac{2^{4}}{3^{4}}=\dfrac{1120}{27}$

Solution

Given expansion is $(a^{3}+b a)^{28}$.

$\because \quad n=28\ $ [even]

$\therefore \quad$ Middle term $=(\dfrac{28}{2}+1)^ {th} $ term $=15^ {th} $ term

$\therefore \quad T _ {15}=T _ {14+1}$

$={ }^{28} C _ {14}(a^{3})^{28-14}(b a)^{14}$

$={ }^{28} C _ {14} a^{42} b^{14} a^{14}$

$={ }^{28} C _ {14} a^{56} b^{14}$

Solution

Given expansion is $(1+x)^{p+q}$.

$\therefore \quad$ Coefficient of $x^{p}={ }^{p+q} C _ {p}$

and coefficient of $x^{q}={ }^{p+q} C _ {q}$

$\therefore \quad \dfrac{{ }^{p+q} C _ {p}}{{ }^{p+q} C _ {q}}=\dfrac{{ }^{p+q} C _ {p}}{{ }^{p+q} C _ {p}}=1: 1$

Solution

Given expansion is $\left(\sqrt{\dfrac{x}{3}}+{\dfrac{3}{2 x^{2}}}\right)^{10}$.

Let the constant term be $T _ {r+1}$.

Then,

$ \begin{aligned} T _ {r+1} & ={ }^{10} C _ {r} \left(\sqrt{\dfrac{x}{3}}\right)^{10-r}{\left(\dfrac{3}{2 x^{2}}\right)}^{r} \\ \\ & ={ }^{10} C _ {r} \cdot x^{\left(\dfrac{10-r}{2}\right)} \cdot 3^{\left(\dfrac{-10+r}{2}\right)} \cdot 3^{r} \cdot 2^{-r} \cdot x^{-2 r} \\ \\ & ={ }^{10} C _ {r} x^{\left(\dfrac{10-5 r}{2}\right)} 3^{\left(\dfrac{-10+3 r}{2}\right)} 2^{-r} \end{aligned} $

For constant term, $10-5 r=0 \Rightarrow r=2$

Hence, third term is independent of $x$.

Solution

Let

$ \begin{aligned} 25^{15} & =(26-1)^{15} \\ \\ & ={ }^{15} C _ 0 26^{15}-{ }^{15} C _ 1 26^{14}+\ldots-{ }^{15} C _ {15} \\ \\ & ={ }^{15} C _ 0 26^{15}-{ }^{15} C _ 1 26^{14}+\ldots-1-13+13 \\ \\ & ={ }^{15} C _ 0 26^{15}-{ }^{15} C _ 1 26^{14}+\ldots-13+12 \end{aligned} $

It is clear that, when $25^{15}$ is divided by 13 , then remainder will be 12 .

Solution

False

Given series

$ \begin{aligned} & =\displaystyle \sum _ {r=0}^{10}{ }^{20} C _ {r}={ }^{20} C _ 0+{ }^{20} C _ 1+{ }^{20} C _ 2+\ldots+{ }^{20} C _ {10} \\ \\ & ={ }^{20} C _ 0+{ }^{20} C _ 1+\ldots+{ }^{20} C _ {10}+{ }^{20} C _ {11}+\ldots{ }^{20} C _ {20}-({ }^{20} C _ {11}+\ldots+{ }^{20} C _ {20}) \\ \\ & =2^{20}-({ }^{20} C _ {11}+\ldots+{ }^{20} C _ {20}) \end{aligned} $

Hence, the given statement is false.

Solution

True

Given expression $=7^{9}+9^{7}=(1+8)^{7}-(1-8)^{9}$

$ \begin{aligned} & =({ }^{7} C _ 0+{ }^{7} C _ 1 8+{ }^{7} C _ 2 8^{2}+\ldots+{ }^{7} C _ 7 8^{7})-({ }^{9} C _ 0-{ }^{9} C _ 1 8+{ }^{9} C _ 2 8^{2} \ldots-{ }^{9} C _ 9 8^{9}) \\ \\ & =(1+7 \times 8+21 \times 8^{2}+\ldots)-(1-9 \times 8+36 \times 8^{2}+\ldots-8^{9}) \\ \\ & =(7 \times 8+9 \times 8)+(21 \times 8^{2}-36 \times 8^{2})+\ldots \\ \\ & =2 \times 64+(21-36) 64+\ldots \end{aligned} $

which is divisible by 64

Hence, the statement is true.

Solution

False

Given expansion is $[(2 x+y^{3})^{4}]^{7}=(2 x+y^{3})^{28}$.

Since, this expansion has 29 terms.

So, the given statement is false.

Solution

False

Here, the Binomial expansion is $(1+x)^{2 n-1}$.

Since, this expansion has two middle term i.e., $\left(\dfrac{2 n-1+1}{2}\right)^{th}$ term and $(\dfrac{2 n-1+1}{2}+1)^{th}$ term i.e., $ n ^{th}$ term and $(n+1)^{th}$ term.

$ \begin{aligned} \therefore \quad \text { Coefficient of } n \text {th term } & ={ }^{2 n-1} C _ {n-1} \\ \\ \text { Coefficient of }(n+1) \text { th term } & ={ }^{2 n-1} C _ {n} \\ \\ \text { Sum of coefficients } & ={ }^{2 n-1} C _ {n-1}+{ }^{2 n-1} C _ {n} \\ \\ & ={ }^{2 n-1+1} C _ {n}={ }^{2 n} C _ {n} \quad[\because{ }^{n} C _ {r}+{ }^{n} C _ {r-1}={ }^{n+1} C _ {r}] \end{aligned} $

Solution

True

Given that, $\quad 3^{400}=9^{200}=(10-1)^{200}$

$ \begin{matrix} \Rightarrow & (10-1)^{200}={ }^{200} C _ 0 10^{200}-{ }^{200} C _ 1 10^{199}+\ldots-{ }^{200} C _ {199} 10^{1}+{ }^{200} C _ {200} 1^{200} \\ \\ \Rightarrow & (10-1)^{200}=10^{200}-200 \times 10^{199}+\ldots-10 \times 200+1 \end{matrix} $

So, it is clear that the last two digits are 01 .

Solution

False

Given Binomial expansion is $\left(x-{\dfrac{1}{x^{2}}}\right)^{2 n}$.

Let $\quad T _ {r+1}$ term is independent of $x$.

Then,

$ \begin{aligned} T _ {r+1} & ={ }^{2 n} C _ {r}(x)^{2 n-r}-{\left(\dfrac{1}{x^{2}}\right)}^{r} \\ \\ & ={ }^{2 n} C _ {r} x^{2 n-r}(-1)^{r} x^{-2 r}={ }^{2 n} C _ {r} x^{2 n-3 r}(-1)^{r} \end{aligned} $

For independent of $x$,

$ \quad \begin{aligned} 2 n-3 r & =0 \\ \\ \therefore r & =\dfrac{2 n}{3}, \end{aligned} $

which is not a integer.

So, the given expansion is not possible.

Solution

False

We know that, the number of terms in the expansion of $(a+b)^{n}$, where $n \in N$, is one more than the power $n$.