1. Bino-geometric series

${ }^{\mathrm{n}} \mathrm{C} _{0}+{ }^{\mathrm{n}} \mathrm{C} _{1} \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C} _{2} \mathrm{x}^{2}………$ $+{ }^{n} C _{n} x^{n}=(1+x)^{n}$

eg. ${ }^{\mathrm{n}} \mathrm{C} _{0}+{ }^{\mathrm{n}} \mathrm{C} _{1} \cdot 3+{ }^{\mathrm{n}} \mathrm{C} _{2} \cdot 3^{2}+$ $…….+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}} 3^{\mathrm{n}}=(1+3)^{\mathrm{n}}=4^{\mathrm{n}}$

2. Bino-arithmetic series

$ a^{n} C _{o}+(a+d){ }^{n} C _{1}+(a+2 d)^{n} C _{2}+\ldots \ldots \ldots \ldots \ldots \ldots+(a+n d)^{n} C _{n} $

This series is the sum of the products of corresponding terms of

${ }^{\mathrm{n}} \mathrm{C} _{0},{ }^{\mathrm{n}} \mathrm{C} _{1},{ }^{\mathrm{n}} \mathrm{C} _{2}………$ ${ }^{n} C _{n}$ (binomial coefficients) and $a, a+d, a+2 d…….,a+nd$ (arithmetic progression)

Such series can be solved either by

(i). eliminating $\mathrm{r}$ in the multiplier of binomial coefficient from the $(\mathrm{r}+1)^{\mathrm{th}}$ terms of the series (i.e. using $\mathrm{r}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}=\mathrm{n}^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}$ ) or

(ii). Differentiating the expansion of $\mathrm{x}^{\mathrm{a}}\left(1+\mathrm{x}^{\mathrm{d}}\right)^{\mathrm{n}}$ or (If product of two or more numericals occur, then differentiate again and again till we get the desired result)

eg. Prove that ${ }^{n} C _{0}+2{ }^{n} C _{1}+3{ }^{n} C _{2}+………$$+(n+1)^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}=(\mathrm{n}+2) \cdot 2^{\mathrm{n}-1}$

${ }^{n} C _{0}+2 \cdot{ }^{n} C _{1}+\ldots . .+(n+1){ }^{n} C _{n}$

$=\sum _{\mathrm{r}=0}^{\mathrm{n}}(\mathrm{r}+1){ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}$

$=\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}+\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}$

$=\sum _{\mathrm{r}=1}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}}{ }^{\mathrm{n}-1} \mathrm{C} _{\mathrm{r}-1}+\sum _{\mathrm{r}=1}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}$

$=n \cdot 2^{\mathrm{n}-1}+2^{\mathrm{n}}=(\mathrm{n}+2) 2^{\mathrm{n}-1}$

OR

Consider the expansion

$ { }^{\mathrm{n}} \mathrm{C} _{0}+{ }^{\mathrm{n}} \mathrm{C} _{1} \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C} _{2} \mathrm{x}^{2}+\ldots \ldots \ldots . .+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}=(1+\mathrm{x})^{\mathrm{n}} $

Multiply by $\mathrm{x}$

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

Differentiate w.r.t.x $ { }^n C_0+{ }^n C_1 2 x+{ }^n C_2 3 x^2+\ldots \ldots \ldots . .+{ }^n C_n(n+1) x^n=x n(1+x)^{n-1}+(1+x)^n $

Put $x=1$ $ { }^n C_0+2^n C_1+3^n C_2+\ldots \ldots \ldots \ldots+(n+1)^n C_n=n 2^{n-1}+2^n=(n+2) 2^{n-1} $

3. Bino-harmonic series

$\frac{{ }^{n} C _{0}}{a}+\frac{{ }^{n} C _{1}}{a+d}+\frac{{ }^{n} C _{2}}{a+2 d}+\ldots \ldots . .+\frac{{ }^{n} C _{n}}{a+n d}$

This series is the sum of the products of corresponding terms of

${ }^{\mathrm{n}} \mathrm{C} _{0},{ }^{\mathrm{n}} \mathrm{C} _{1},{ }^{\mathrm{n}} \mathrm{C} _{2}$, ${ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}$ (binomial coefficients) and

$\frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2 d}, \ldots \ldots \ldots \ldots \ldots . . . \frac{1}{a+n d}$ (harmonic progression)

Such seris can be solved either by

(i). eliminating $\mathrm{r}$ in the multiplier of binomial coefficient from the $(\mathrm{r}+1)^{\mathrm{th}}$ term of the series (ie using $\frac{1}{\mathrm{r}+1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}=\frac{1}{\mathrm{n}+1}{ }^{\mathrm{n}+1} \mathrm{C} _{\mathrm{r}+1}$ ) or

(ii). integrating suitable expansion

Note

(i). If the sum contains $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}……$ $\mathrm{C} _{\mathrm{n}}$ are all positive signs, integrate between limits 0 to 1

(ii). If the sum contains alternate signs (i.e.+ & -) then integrate between limits -1 to 0

(iii). If the sum contains odd coefficients (i.e. $\mathrm{C} _{\mathrm{o}}, \mathrm{C} _{2}, \mathrm{C} _{4}, \ldots \ldots$. ) then integrate between $-1 to +1 .$

(iv). If the sum contains even coefficient (i.e. $\mathrm{C} _{1}, \mathrm{C} _{3}, \mathrm{C} _{5}, \ldots \ldots \ldots$ ) the find the difference between (i) & (iii). and then divide by 2

(v). If in denominator of binomial coefficient is product of two numericals then integrate two times first time take limits between 0 to $\mathrm{x}$ and second time take suitable limits

eg: prove that $\frac{{ }^{n} C _{0}}{1}+\frac{{ }^{n} C _{1}}{2}+\frac{{ }^{n} C _{2}}{3}+\ldots \ldots \ldots \ldots \ldots . .+\frac{{ }^{n} C _{n}}{n+1}=\frac{2^{n+1}-1}{n+1}$

${ }^{n} C _{0}+\frac{{ }^{n} C _{1}}{2}+\frac{{ }^{n} C _{2}}{3}+\ldots \ldots \ldots \ldots . .+\frac{{ }^{n} C _{n}}{n+1}=\sum _{r=0}^{n} \frac{{ }^{n} C _{r}}{r+1}$

$=\frac{1}{\mathrm{n}+1} \sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}}{\mathrm{r}+1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}$

$=\frac{1}{\mathrm{n}+1} \sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}+1} \mathrm{C} _{\mathrm{r}+1}$

$=\frac{1}{n+1} \sum_{r=0}^n{ }^{n+1} C_{r+1}$

$=\frac{1}{\mathrm{n}+1}\left(2^{\mathrm{n}+1}-1\right)$

OR

Consider the expansion

$ (1+x)^{n}={ }^{n} C _{0}+{ }^{n} C _{1} x+{ }^{n} C _{2} x^{2}+\ldots \ldots \ldots . .+{ }^{n} C _{n} x^{n} $

Integrate between limit to 0 to 1

$ \begin{aligned} & {\left[\frac{(1+x)^{n+1}}{n+1}\right] _{0}^{1}=\left[{ }^{n} C _{0} x+{ }^{n} C _{1} \frac{x^{2}}{2}+{ }^{n} C _{2} \frac{x^{3}}{3}+\ldots \ldots .+\frac{{ }^{n} C _{n} x^{n+1}}{n+1}\right] _{0}^{1}} \\ & \left(\frac{2^{n+1}}{n+1}-\frac{1}{n+1}\right)={ }^{n} C _{0}+\frac{{ }^{n} C _{1}}{2}+\frac{{ }^{n} C _{2}}{3}+\ldots \ldots \ldots .+\frac{{ }^{n} C _{n}}{n+1} \end{aligned} $

$ \therefore{ }^{\mathrm{n}} \mathrm{C} _{0}+\frac{{ }^{\mathrm{n}} \mathrm{C} _{1}}{2}+\frac{{ }^{\mathrm{n}} \mathrm{C} _{2}}{3}+\ldots \ldots \ldots+\frac{{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}}{\mathrm{n}+1}=\frac{1}{\mathrm{n}+1}\left(2^{\mathrm{n}+1}-1\right) $

4. Bino-binomial series

• ${ }^{\mathrm{n}} \mathrm{C} _{0}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C} _{1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}+1}+{ }^{\mathrm{n}} \mathrm{C} _{2}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}+2}+…………$ $+{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}-\mathrm{r}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}} \quad$ or

• ${ }^{\mathrm{m}} \mathrm{C} _{0}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}+{ }^{\mathrm{m}} \mathrm{C} _{1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}-1}+{ }^{\mathrm{m}} \mathrm{C} _{2}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}-2}+\ldots \ldots \ldots . .+{ }^{\mathrm{m}} \mathrm{C} _{\mathrm{r}}{ }^{\mathrm{n}} \mathrm{C} _{0}$

Such series can be solved by multiplying two expansions, one involving the first factors as coefficient and the other involving the second factors as coefficients and finally equating coefficients of a suitable power of $\mathrm{x}$ on both sides.

Prove That

${ }^{\mathrm{n}-1} \mathrm{C} _{0}{ }^{\mathrm{n}} \mathrm{C} _{1}+{ }^{\mathrm{n-}-1} \mathrm{C} _{1} \mathrm{n} _{2} \mathrm{C} _{2}+{ }^{\mathrm{n}-1} \mathrm{C} _{2}{ }^{\mathrm{n}} \mathrm{C} _{3}+\ldots \ldots \ldots . .+{ }^{+\mathrm{n-}-1} \mathrm{C} _{\mathrm{n}-1}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}={ }^{2 \mathrm{n}-1} \mathrm{C} _{\mathrm{n}-1}$

We have

$\left({ }^{n} C _{0} x^{n+}+{ } _{1} C _{1} x^{n-1}+{ }^{n} C _{2} x^{n-2}+\ldots \ldots .+{ }^{n} C _{n-1} x^{+n} C _{n}\right)\left({ }^{n-1} C _{0}+{ }^{n-1} C _{1} x^{n-1} C _{2} x^{2}+\ldots \ldots \ldots . .+{ }^{n-1} C^{n-1} x^{n-1}\right)=(1+x)^{n}$ $(1+\mathrm{x})^{\mathrm{n}-1}$

$\Rightarrow\left({ }^{n} C _{0} x^{n}+{ }^{n} C _{1} x^{n-1}+{ }^{n} C _{2} x^{n-2}+\ldots .+{ }^{n} C _{n-1} x+{ }^{n} C _{n}\right)\left({ }^{n-1} C _{0}+{ }^{n-1} C _{1} x+{ }^{n-1} C _{2} x^{2}+\ldots . .+{ }^{n-1} C _{n-1} x^{n-1}\right)=(1+x)^{2 n-1}$ Equate the coefficients of $x^{n-1}$ on both sides, ${ }^{n-1} \mathrm{C} _{0}{ }^{n} C _{1}+{ }^{n-1} C _{1}{ }^{n} C _{2}+$ ${ }^{n-1} C _{n-1}{ }^{n} C _{n}={ }^{2 n-1} C _{n-1}$

Note : For the sake of convenience, the coefficients ${ }^{n} \mathrm{C} _{0},{ }^{n} \mathrm{C} _{1}, \ldots \ldots . .{ }^{n} \mathrm{C} _{\mathrm{r}}, \ldots . .{ }^{n} \mathrm{C} _{\mathrm{n}}$ are usually denoted by $\mathrm{C} _{0}, \mathrm{C} _{1}$, $\mathrm{C} _{\mathrm{r}}$ . $\mathrm{C} _{\mathrm{n}}$ respectively

Use of complex numbers in Binomial Theorem

We know $(\cos \theta+\sin \theta)^{\mathrm{n}}=\cos \theta \theta+\sin n \theta$.

Expand and the binomial and then equating the real and imaginary parts, we get

$\cos \theta=\cos ^{n} \theta-{ }^{n} C _{2} \cos ^{n-2} \theta \sin ^{2} \theta+{ }^{n} C _{4} \cos ^{n-4} \theta \sin ^{4} \theta+…..$

$\sin n \theta={ }^{n} C _{1} \cos ^{n-1} \theta \sin \theta-{ }^{n} C _{3} \cos ^{n-3} \theta \sin ^{3} \theta+{ }^{n} C _{5} \cos ^{n-5} \theta \sin ^{5} \theta+……..$

$\Rightarrow \operatorname{tann} \theta=\frac{{ }^{n} C _{1} \tan \theta-{ }^{n} C _{1} \tan ^{3} \theta+{ }^{n} C _{5} \tan ^{5} \theta \ldots \ldots \ldots \ldots . . . . .}{1-{ }^{n} C _{2} \tan ^{2} \theta+{ }^{n} C _{4} \tan ^{4} \theta-{ }^{n} C _{6} \tan ^{6} \theta+\ldots . .}$

Solved examples

1. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \mathrm{C} _{3}, \ldots \ldots \ldots . . \mathrm{C} _{\mathrm{n}-1}, \mathrm{C} _{\mathrm{n}}$ denote the binomial coefficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that

$ \begin{aligned} \mathrm{C} _{0} \mathrm{C} _{1}+\mathrm{C} _{1} \mathrm{C} _{2}+\mathrm{C} _{2} \mathrm{C} _{3}+\ldots \ldots . . & +\mathrm{C} _{\mathrm{n}-1} \mathrm{C} _{n} \\ & =\frac{(2 \mathrm{n}) !}{(\mathrm{n}-1) !(n+1) !} \\ & =\frac{1.3 .5(2 \mathrm{n}-1)}{(\mathrm{n}+1) !} \cdot \mathrm{n} .2^{\mathrm{n}} \end{aligned} $

Show Answer

Solution :

Using binomial expansion, we have

$ (1+x)^{n}=C _{0}+C _{1} x+C _{2} x^{2}+\ldots \ldots . .+C _{r} x^{r}+\ldots .+C _{n} x^{n} \ldots \ldots . .(A) \text { and } $

$ (1+x)^{n}=C _{0} x^{n}+C _{1} x^{n-1}+C _{2} x^{n-2}+\ldots \ldots \ldots+C _{r} x^{n-r}+\ldots \ldots \ldots \ldots .+\ldots \ldots .+C _{n-1} x+C _{n} \ldots \ldots(B) $

Multiplying (A) and (B), we get

$(1+\mathrm{x})^{2 \mathrm{n}}$

$=\left(\mathrm{C} _{0}+\mathrm{C} _{1} \mathrm{x}+\mathrm{C} _{2} \mathrm{x}^{2}+\ldots . .+\mathrm{C} _{\mathrm{r}} \mathrm{r}^{+}+\ldots .+\mathrm{C} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right) \times\left(\mathrm{C} _{0} \mathrm{x}^{\mathrm{n}}+\mathrm{C} _{1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{C} _{2} \mathrm{x}^{\mathrm{n}-2}+\ldots .+\mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}}+\ldots \ldots+\mathrm{C} _{\mathrm{n}-1} \mathrm{x}+\mathrm{C} _{\mathrm{n}}\right)$

or

$ \left(C _{0}+C _{1} x+C _{2} x^{2}+\ldots \ldots .+C _{r} x^{r}+\ldots \ldots+C _{n} x^{n}\right)\left(C _{0} x^{n}+C _{1} x^{n-1}+C 2^{n-2}+\ldots \ldots+C _{r} x^{n-r}+\ldots \ldots+C _{n-1}\right.$ $\left.x+C_n\right)=(1+x)^{2 n} \ldots \ldots \ldots \ldots \ldots \ldots . . . . . .(C)$

Equating the coefficients of $\mathrm{x}^{\mathrm{n-1}}$ on both sides of $(\mathrm{C})$, we get

$ \begin{aligned} & C _{0} C _{1}+C _{1} C _{2}+\ldots \ldots+C _{n-1} C _{n}={ }^{2 n} C _{n-1} \\ & \Rightarrow C _{0} C _{1}+C _{1} C _{2}+\ldots \ldots+C _{n-1} C _{n}={ }^{2 n} C _{n-1}=\frac{(2 n) !}{(n+1) !(n-1) !} \end{aligned} $

Now,

$ \begin{aligned} & \frac{(2 n) !}{(\mathrm{n}+1) !(\mathrm{n}-1) !} \\ & =\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \ldots \ldots \ldots .(2 \mathrm{n}-2)(2 \mathrm{n}-1)(2 \mathrm{n})}{(\mathrm{n}+1) !(\mathrm{n}-1) !} \\ & =\frac{\{1.3 \cdot 5 \ldots \ldots .(2 \mathrm{n}-1)\}\{2 \cdot 4 \cdot 6 \ldots \ldots . .2 \mathrm{n}\}}{(\mathrm{n}+1) !(\mathrm{n}-1) !} \\ & =\frac{\{1 \cdot 3.5 \ldots \ldots .(2 \mathrm{n}-1)\} .2^{\mathrm{n}} \cdot \mathrm{n} !}{(\mathrm{n}+1) !(\mathrm{n}-1) !} \\ & =\frac{1.3 \cdot 5 \ldots \ldots .(2 \mathrm{n}-1)}{(\mathrm{n}+1) !} \frac{2^{\mathrm{n}} \cdot \mathrm{n} \cdot(\mathrm{n}-1) !}{(\mathrm{n}-1) !} \\ & =\frac{1.3 \cdot 5 \ldots .(2 \mathrm{n}-1)}{(\mathrm{n}+1) !} n \cdot 2^{\mathrm{n}} \end{aligned} $

Hence,

$ \begin{aligned} C _{0} C _{1}+C _{1} C _{2}+C _{2} C _{3}+\ldots \ldots \ldots+C _{n-1} C _{n} & =\frac{(2 n) !}{(n-1) !(n+1) !} \\ & =\frac{1.3 .5 \ldots(2 n-1)^{n} \cdot 2^{n}}{(n+1) !} \end{aligned} $

2. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \mathrm{C} _{3}, \ldots \ldots \ldots . \mathrm{C} _{\mathrm{n}-1}, \mathrm{C} _{\mathrm{n}}$ denote the binomial coefficients in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that

$ \begin{aligned} \mathrm{C} _{0}^{2}+\mathrm{C} _{1}^{2}+\mathrm{C} _{2}^{2}+\ldots \ldots \ldots \mathrm{C} _{\mathrm{n}}^{2} & =\frac{(2 \mathrm{n}) !}{(\mathrm{n} !)^{2}} \\ & =\frac{1.3 .5 \ldots(2 \mathrm{n}-1)}{\mathrm{n} !} 2^{\mathrm{n}} \end{aligned} $

Show Answer

Solution :

Using binomial expansion, we have

$\begin{aligned} & (1+x)^n=C_0+C_1 x+C_2 x^2+\ldots \ldots .+C_r x^r+\ldots .+C_n x^n \ldots \ldots . .(A) \text { and } \\ & (1+x)^n=C_0 x^n+C_1 x^{n-1}+C_2 x^{n-2}+\ldots \ldots \ldots+C_r x^{-r}+\ldots \ldots \ldots \ldots . . \ldots \ldots+C_{n-1} x+C_n \ldots \ldots .(B)\end{aligned}$

Multiplying (A) and (B), we get

$(1+\mathrm{x})^{2 \mathrm{n}}$

$=\left(\mathrm{C} _{0}+\mathrm{C} _{1} \mathrm{x}+\mathrm{C} _{2} \mathrm{x}^{2}+\ldots .+\mathrm{C} _{\mathrm{r}} \mathrm{r}^{\mathrm{r}}+\ldots . .+\mathrm{C} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right) \times\left(\mathrm{C} _{0} \mathrm{x}^{\mathrm{n}}+\mathrm{C} _{1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{C} _{2} \mathrm{x}^{\mathrm{n}-2}+\ldots .+\mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}}+\ldots \ldots .+\mathrm{C} _{\mathrm{n}-1} \mathrm{x}+\mathrm{C} _{\mathrm{n}}\right)$ or $\quad\left(\mathrm{C} _{0}+\mathrm{C} _{1} \mathrm{x}+\mathrm{C} _{2} \mathrm{x}^{2}+\right.$ $\mathrm{x}+\mathrm{C})=(1+\mathrm{x})^{2 \mathrm{n}}$ $+\mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{r}}+$ $\left.+\mathrm{C} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right)\left(\mathrm{C} _{0} \mathrm{x}^{\mathrm{n}}+\mathrm{C} _{1} \mathrm{x}^{\mathrm{n}-1}+\mathrm{C}^{\mathrm{n}-2}+\ldots\right.$ $+\mathrm{C} _{\mathrm{r}} \mathrm{x}^{\mathrm{n}-\mathrm{r}}+\ldots \ldots+\mathrm{C} _{\mathrm{n}-1}$$\left.x+C_n\right)=(1+x)^{2 n} \ldots \ldots \ldots \ldots \ldots \ldots . . . . . . .(C)$

Equating the coefficients of $\mathrm{x}^{\mathrm{n}}$ on both sides of $(\mathrm{C})$, we get

$ C_0{}^2+C_1{ }^2+C_2{}^2+\ldots \ldots.C_{\mathrm{n}}{ }^2={ }^{2 \mathrm{n}} \mathrm{C}_{\mathrm{n}} $

$\rarr C_0^2 + C_1^2 + C_2^2+……..C_n^2=\frac{(2n)!}{n!n!}$

Now

$ \begin{aligned} & \frac{(2 \mathrm{n}) !}{\mathrm{n} ! \mathrm{n} !}=\frac{1.2 .3 .4 .5 \ldots \ldots \ldots .(2 \mathrm{n}-2)(2 \mathrm{n}-1)(2 \mathrm{n})}{\mathrm{n} ! \mathrm{n} !} \\ & =\frac{\{1.3 .5 \ldots \ldots .(2 \mathrm{n}-1)\}\{2.4 .6 \ldots \ldots .(2 \mathrm{n}-2)(2 \mathrm{n})\}}{\mathrm{n} ! \mathrm{n} !} \\ & =\frac{\{1.3 .5 \ldots \ldots .(2 \mathrm{n}-1)\} \times 2^{\mathrm{n}} \times\{1.2 .3 \ldots \ldots \ldots(\mathrm{n}-1) \cdot \mathrm{n}\}}{\mathrm{n} ! \mathrm{n} !} \\ & =\frac{\{1.3 .5 \ldots \ldots .(2 \mathrm{n}-1)\} 2^{\mathrm{n}} \mathrm{n} !}{\mathrm{n} ! \mathrm{n} !} \\ & =\frac{1.3 .5 \ldots \ldots .(2 \mathrm{n}-1)}{\mathrm{n} !} 2^{\mathrm{n}} \\ & \text { Hence, } \mathrm{C} _{0}{ }^{2}+\mathrm{C} _{1}{ }^{2}+\mathrm{C} _{2}{ }^{2}+\ldots \ldots . \mathrm{C} _{\mathrm{n}}{ }^{2}=\frac{(2 \mathrm{n}) !}{\mathrm{n} ! \mathrm{n} !} \\ & \quad=\frac{1.3 .5 \ldots \ldots .(2 \mathrm{n}-1)}{\mathrm{n} !} 2^{\mathrm{n}} \end{aligned} $

3. If $\mathrm{C} _{0}, \mathrm{C} _{1}, \mathrm{C} _{2}, \ldots \ldots . \mathrm{C} _{\mathrm{n}}$ denote the binomial fulfillments in the expansion of $(1+\mathrm{x})^{\mathrm{n}}$, prove that ; $\mathrm{C} _{0}{ }^{2}-\mathrm{C} _{1}{ }^{2}+\mathrm{C} _{2}{ }^{2}-\mathrm{C} _{3}{ }^{2}+\ldots \ldots .+(-1)^{\mathrm{n}} \mathrm{C} _{\mathrm{n}}{ }^{2}$

$ = \begin{cases}0 & \text {,if } n \text { is odd } \\ (-1)^{n / 2} \cdot C _{n / 2} & \text {,if } n \text { is even }\end{cases} $

4. If $(1+x)^{n}=C _{0}+C _{1} x+C _{2} x^{2}+\ldots .+C _{n} x^{n}$ prove that $\sum _{r=0}^{n} \sum _{s=0}^{n}\left(C _{r}+C _{s}\right)=(n+1)$

5. If $(1+x)^{n}=C _{0}+C _{1} x+C _{2} x^{2}+\ldots+C _{n} x^{n}$ prove that $\sum _{r=0}^{n} \sum _{s=0}^{n} C _{r} C _{s}=2^{2 n}$

6. If $(1+\mathrm{x})^{\mathrm{n}}=\mathrm{C} _{0}+\mathrm{C} _{1} \mathrm{x}+\mathrm{C} _{2} \mathrm{x}^{2}+\ldots+\mathrm{C} _{\mathrm{n}} \mathrm{x}^{\mathrm{n}}$ prove that

$ \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)=\mathrm{n} \cdot 2^{\mathrm{n}} $

Show Answer

Solution: We have,

$\sum _{\mathrm{r}=0}^{\mathrm{n}} \sum _{\mathrm{s}=0}^{\mathrm{n}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)=\sum _{\mathrm{r}=0}^{\mathrm{n}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)+2 \sum _{0 \leq \mathrm{r} r \mathrm{~s} \leq \mathrm{n}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)$

$\Rightarrow(\mathrm{n}+1) 2^{\mathrm{n}+1}=2\left(\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\right)+2 \sum _{0 \leq \mathrm{r}<\leq \leq \mathrm{n}} \sum _{\mathrm{r}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)$

$\Rightarrow(\mathrm{n}+1) 2^{\mathrm{n}+1}=2 \cdot 2^{\mathrm{n}}+2 \sum _{0 \leq \mathrm{r}<\leq \leq \mathrm{n}} \sum _{\mathrm{r}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)$

$\Rightarrow \mathrm{n} .2^{\mathrm{n}+1}=2 \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)$

$\Rightarrow \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}}\left(\mathrm{C} _{\mathrm{r}}+\mathrm{C} _{\mathrm{s}}\right)=\mathrm{n} .2^{\mathrm{n}}$

7. If $(1+x)^{n}=C _{0}+C _{1} x+C _{2} x^{2}+\ldots+C _{n} x^{n}$ prove that $\sum _{0 \leq r<s \leq n} C _{r} C _{s}=\frac{1}{2}\left(2^{2 n _2 n} C _{n}\right)$

Show Answer

Solution: We have,

$ \begin{aligned} & \sum _{\mathrm{r}=0}^{\mathrm{n}} \sum _{\mathrm{s}=0}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{C} _{\mathrm{s}}=\left(\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}^{2}\right)+2 \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{C} _{\mathrm{s}} \\ & \Rightarrow 2^{2 \mathrm{n}}={ }^{2 \mathrm{n}} \mathrm{C} _{\mathrm{n}}+2 \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{C} _{\mathrm{s}} \\ & \Rightarrow \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{C} _{\mathrm{s}}=\frac{1}{2}\left[2^{2 \mathrm{n}-2 \mathrm{n}} \mathrm{C} _{\mathrm{n}}\right] \end{aligned} $

ALITER We have,

$ \begin{aligned} & \left(\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\right)^{2}=\left(\sum _{\mathrm{r}=0}^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}^{2}\right)+2\left(\sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{C} _{\mathrm{s}}\right) \\ & \Rightarrow\left(2^{\mathrm{n}}\right)^{2}={ }^{2 \mathrm{n}} \mathrm{C} _{\mathrm{n}}+2\left(\sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{C} _{\mathrm{s}}\right) \\ & \Rightarrow \sum _{0 \leq \mathrm{r}<\mathrm{s} \leq \mathrm{n}} \sum _{\mathrm{r}} \mathrm{C} _{\mathrm{s}}=\frac{1}{2}\left[2^{2 \mathrm{n}-2 \mathrm{n}} \mathrm{C} _{\mathrm{n}}\right] \end{aligned} $

Practice questions

1. If $n$ is an integer between 0 and 21 , then the minimum value of $n$ ! (21-n)! is attained for $n=$

(a) 1

(b) 10

(c) 12

(d) 20

2. ${ }^{404} \mathrm{C} _{4}-{ }^{4} \mathrm{C} _{1}{ }^{303} \mathrm{C} _{4}+{ }^{4} \mathrm{C} _{2}{ }^{202} \mathrm{C} _{4}{ }^{-4} \mathrm{C} _{3}{ }^{101} \mathrm{C} _{4}$ is equal to

(a) $(401)^{4}$

(b) $(101)4$

(c) $0$

(d) (201)4

3. If $\left(3+x^{2008}+x^{2009}\right)^{2010}=a _{0}+a _{1} x+a _{2} x^{2}+\ldots \ldots .+a _{n} x^{n}$, then the value of $a _{0}-\frac{1}{2} a _{1}-\frac{1}{2} a _{2}+a _{3}-\frac{1}{2} a _{4}-$ $\frac{1}{2} \mathrm{a} _{5}+\mathrm{a} _{6} \ldots \ldots$. is

(a) $3^{2010}$

(b) $1$

(c) $2^{2010}$

(d) None of these

4. $\left\{\left({ }^{n} \mathrm{C} _{0}+{ }^{\mathrm{n}} \mathrm{C} _{3}+\ldots \ldots \ldots . .\right)-\frac{1}{2}\left({ }^{\mathrm{n}} \mathrm{C} _{1}+{ }^{\mathrm{n}} \mathrm{C} _{2}+{ }^{\mathrm{n}} \mathrm{C} _{4}+{ }^{\mathrm{n}} \mathrm{C} _{5} \ldots \ldots \ldots .\right)\right\}^{2}+\frac{3}{4}\left({ }^{3} \mathrm{C} _{1}-{ }^{\mathrm{n}} \mathrm{C} _{2}+{ }^{\mathrm{n}} \mathrm{C} _{4}-{ }^{\mathrm{n}} \mathrm{C} _{5}+\ldots \ldots .\right)^{2}=$

(a) 3

(b) 4

(c) 2

(d) 1

5. Value of $\sum _{\mathrm{r}=0}^{20} \mathrm{r}(20-\mathrm{r})\left({ }^{20} \mathrm{C}\right){ }^{2}$ is equal to

(a) $400{ }^{39} \mathrm{C} _{20}$

(b) $400{ }^{40} \mathrm{C} _{19}$

(c) $400{ }^{39} \mathrm{C} _{19}$

(d) $400{ }^{38} \mathrm{C} _{20}$

6. If for $z$ as real or complex,$\left(1+\mathrm{z}^{2}+\mathrm{z}^{4}\right)^{8}=\mathrm{C} _{0}+\mathrm{C} _{1} \mathrm{z}^{2}+\mathrm{C} _{2} \mathrm{Z}^{4}+$. $+\mathrm{C} _{16} \mathrm{Z}^{32}$, then

(a) $\mathrm{C} _{0}-\mathrm{C} _{1}+\mathrm{C} _{2}-\mathrm{C} _{3}+$ $+\mathrm{C} _{16}=1$

(c) $\mathrm{C} _{2}+\mathrm{C} _{5}+\mathrm{C} _{8}+\mathrm{C} _{11}+\mathrm{C} _{14}=3^{6}$

(b) $\mathrm{C} _{0}+\mathrm{C} _{3}+\mathrm{C} _{6}+\mathrm{C} _{9}+\mathrm{C} _{15}=3^{7}$

(d) $\mathrm{C} _{1}+\mathrm{C} _{4}+\mathrm{C} _{7}+\mathrm{C} _{10}+\mathrm{C} _{13}+\mathrm{C} _{16}=3^{7}$

7. Read the passage and answer the following questions

Any complex number in polar form can be an unpleasing in Euler’s form as $\cos \theta+\sin \theta=\mathrm{e}^{\mathrm{i} \theta}$ which is useful is finding the sum of series $\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}(\cos \theta+\mathrm{i} \sin \theta)^{\mathrm{r}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}(\cos \mathrm{r} \theta+\mathrm{i} \sin \mathrm{r} \theta)$

$=\sum _{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}} \mathrm{e}^{\mathrm{i \theta r}}=\left(1+\mathrm{e}^{\mathrm{i} \theta}\right)^{\mathrm{n}}$

Also we know that the sum of binomial series does not change if $\mathrm{r}$ is replaced by $\mathrm{n}-\mathrm{r}$.

(i). Value of $\sum _{\mathrm{r}=0}^{100}{ }^{100} \mathrm{C} _{\mathrm{r}} \sin (\mathrm{rx} 0)$ is …..

(a) $2^{100} \cos \left(\frac{\mathrm{x}}{2}\right) \sin (50 \mathrm{x})$ $2^{100} \sin 50 x \cos \frac{x}{2}$

(c) $2^{101} \cos (50 \mathrm{x}) \sin \left(\frac{\mathrm{x}}{2}\right)$

(d) $2^{101} \sin ^{100} 50 \mathrm{x} \cos 50 \mathrm{x}$

ii. In triangle $\mathrm{ABC}$, the value of $\sum _{\mathrm{r}=0}^{50}{ }^{50} \mathrm{C} _{\mathrm{r}} \mathrm{a}^{\mathrm{r}} \mathrm{b}^{\mathrm{n-r}} \cos (\mathrm{rb}-(50-\mathrm{r}) \mathrm{A})$ is equal to (a,b,c are sides opposite to A, B, C & S in semi perimeter)

(a) $\mathrm{c}^{49}$

(b) $(a+b)^{50}$

(c) $(25-\mathrm{a}-\mathrm{b})^{50}$

(d) None of these

(iii). If $f(x)=\frac{\sum _{r=0}^{50} C _{r}{ }^{50} \sin 2 \mathrm{rx}}{\sum _{\mathrm{r}=0}^{50} \mathrm{C} _{\mathrm{r}} \cos 2 \mathrm{rx}}$, then $\mathrm{f}(\pi / 8)$ is

(a) $1$

(b) $-1$

(c) irrational value

(d) None of these

8. Match the following

9. The coefficient of $\lambda^{n} \mu^{n}$ is the expansion of $(1+\lambda)^{n}(1+\mu)^{n}(\lambda+\mu)^{n}$ is

(a) $\sum _{\mathrm{r}=0}^{\mathrm{n}}\left({ }^{n} \mathrm{C} _{\mathrm{r}}\right)^{2}$

(b) $\sum _{\mathrm{r}=0}^{\mathrm{n}}\left({ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}-2}\right)^{2}$

(c) $\sum _{\mathrm{r}=0}^{\mathrm{n}}\left({ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}+3}\right)^{2}$

(d) $\sum _{\mathrm{r}=0}^{\mathrm{n}}\left({ }^{\mathrm{n}} \mathrm{C} _{\mathrm{r}}\right)^{3}$

10. If $\mathrm{C} _{1}, \mathrm{C} _{2} \ldots \ldots \ldots . \mathrm{C} _{\mathrm{n}}$ are binomial Coefficients, then the value of $\mathrm{C} _{1}{ }^{2}-2 \mathrm{C} _{2}{ }^{2}+3 \mathrm{C} _{3}{ }^{2}-\ldots \ldots \ldots .-2 \mathrm{nC}^{2}{ } _{2 \mathrm{n}}$ is

(a) $\mathrm{n}^{2}$

(b) $(-1)^{\mathrm{n}-1} \mathrm{n}$

(c) $2(-1)^{\mathrm{n}-1} \mathrm{n}^{2 \mathrm{n}-1} \mathrm{C} _{\mathrm{n}}$

(d) $-n^{2}$