Tricks and Shortcuts#

1. Binomial Theorem#

1. Formula:

$$(a+b)^n=\sum_{r=0}^n {}^nC_r a^{n-r}b^{r}$$

• (n) choose (r), where (n\ge r), or,

$${}^n C_r=\frac{n!}{r!(n-r)!}$$

---

2. Binomial Coefficients:#

$${}^nC_r= {}^nC_{n-r} $$

$${}^nC_0 = {}^nC_n=1$$

• If n = 2k, the middle coefficient is

$$= ^{n}C_{k} = \frac {(2k)!} {(k!)^2}$$

---

##3. Summations/Identities

$$\sum_{i=0}^n {}^nC_i = 2^n$$

$$\sum_{i=0}^n {}^nC_i r^i = (1+r)^n$$

$$\sum_{i=0}^n{}^nC_ir^{n-i}=(1+r)^n$$

• Product of two Binomials:

$$ (a+b)(c+d) = ac+ad+bc+bd $$

---

4. Index Rule#

• Sum of indices of (x) and (y) in any term of the expansion of ((x + y)^n) = n

---

5. Properties of the middle term:#

• The middle term(s) of the expansion of $(a + b)^n$ is determined by the value of $n$. If $n$ is even, there is one middle term; if $n$ is odd, there are two middle terms.

when n is even: there will be two middle terms, that are, ((^{n/2}C_{n/2-1}a^{n/2-1}b^{n/2})) and ((^{n/2}C_{n/2}a^{n/2}b^{n/2-1}))

when n is odd: there will be one middle term, i.e., $$^{n}C_{(n+1)/2}a^{(n-1)/2}b^{(n+1)/2}$$

---

6. Last term#

• The last term of the expansion of $(a+b)^n$ = $b^n$

---

7. General Term:#

• The $(r+1)$th term in the expansion of $(a + b)^n$: $$ T_{r+1} = {}^nC_r a^{n-r} b^r$$

---

Practice Problems#

---

1. If the middle term in the expansion of ((x + \frac {a}{x})^n) is 12870, then find n.#

Solution:

Using the middle term property for the binomial expansion, $$T_{\frac {n}{2}+1}= {}^{n} C_{\frac {n} {2}}x^{n-{n \over 2}+1}\left ( \frac{a}{x} \right )^{\frac {n}{2}-1}$$

$$Rightarrow \space 12870 = {}^n C_{ {n \over 2}-1} x^{{n \over 2}+1} a^{ {n \over 2}-1}$$

$$Rightarrow 12870= \frac {n!} {{({n \over 2}-1)!({n \over 2}+1)!} } x^2 a^{n \over 2}-1 $$

Here we can substitute $n = 12$ to satisfy the equation.

Thus, the value of n is 12.

---

2. If the first, third, and sixth terms of a binomial expansion are a, b, and c respectively, then find the 9th term.#

Solution: General term of a binomial expansion:

$$T_{r+1} = {}^nC_r a^{n-r}b^r$$

Given:

• (T_1 = a = {}^nC_0a^n)

• (T_3 = b = {}^nC_2a^{n-2}b^2)

• $T_6 = c = {}^nC_6a^{n-6}b^6$

• (T_9=?)

• From $$T_3 = {}^nC_2a^{n-2}b^2 = b^2$$

$$Rightarrow b= {}^nC_2a^{n-2}b$$

$$ \Rightarrow \space \frac {b^2}{a^{n-2}} = {}^nC_2 = \frac {n(n-1)}{2a^{n-2}}$$

• Similarly, from $$T_6 = {}^nC_5 a^{n-5}b^5= c$$

We obtain,

$$\Rightarrow \space \frac {c^5}{a^{n-5}b^5} = {}^nC_5 = \frac {n(n-1)(n-2)(n-3)(n-4)}{5!}$$

Dividing Equation 2 by Equation 1:

$$\frac {\frac {c^5}{a^{n-5}b^5}} {\frac {b^2}{a^{n-2}}} =\frac {n(n-1)(n-2)(n-3)(n-4)}{5!}\cdot \frac {2}{n(n-1)}$$

$$ \frac {c^5 b^2 a^3} {a^3 b^8} =\frac {2(n-2)(n-3)(n-4)}{5!}$$

$$\Rightarrow \space c =\frac {2(n-2)(n-3)(n-4)}{5!} b = 2\times \frac {(n-2)(n-3)(n-4)}{3\cdot 2!} b\dots (1)$$

Again from Equation 1,

$$ \frac {b^2}{a^{n-2}} = \frac {n(n-1)}{2}a^{2}$$

$$ \Rightarrow \space a^{n-2} =\frac {2b^2}{n(n-1)}$$

$$ \Rightarrow \space a^3 = \frac {2b^3}{n(n-1)}$$

$$ \Rightarrow \space a =\sqrt[3]{\frac {2b^3}{n(n-1)}}$$

Substituting the value of (a) in Equation (1), we get $9^\text{th}$ term,

$$T_9 = {}^nC_8 a^{n-8}b^8 $$ $$= {}^nC_8\left (\sqrt[3]{\frac {2b^3}{n(n-1)}} \right )^{n-8}b^8$$

Substituting the value of ({}^nC_r) from the property

$${}^nC_8 = {}^nC_{n-8} = \frac {n!}{(n-8)!8!} = \frac {n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{8!}$$

$$T_9 =\frac {n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{8!} \left (\sqrt[3]{\frac {2b^3}{n(n-1)}} \right )^{n-8} b^8$$

Simplifying, we obtain

$$T_9 = \frac {(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{3\times 4!} b^5 = \frac {(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)b^5}{72}$$

---

3. Given that the binomial expansion of ((3x^2 – \frac {1}{2x})^n) contains twelve terms. Find the sum of all possible products of the exponents of (x) in any term.#

Solution:

Binomial expansion of $(3x^2 - \frac {1} {2x})^n$ is valid only when $n$ is a positive integer.

$$(3x^2 - \frac {1} {2x})^n = \sum_{r=0}^n {}^nC_r (3x^2)^{n-r} (-\frac {1} {2x})^r$$

$$= \sum_{r=0}^n {}^nC_r 3^{n-r}(-1)^r x^{2n-3r}2^{-r} $$

By the given condition,

$$2n - 2r -r = 11 \Rightarrow r= \frac{13}{3} \notin N$$

Which implies that n must be a fraction (not an integer)! This means that the given expansion doesn’t exists as a finite binomial expansion. The expansion