Binomial Theorem Tricks and Shortcuts

Tricks and Shortcuts

1. Binomial Theorem

1. Formula:

$$(a+b{)}^n=\sum _{r=0}^n{}^n{C}_r{a}^{n-r}{b}^r$$

• (n) choose (r), where (n\ge r), or,

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

2. Binomial Coefficients:

$${}^n{C}_r={}^n{C}_{n-r}$$

$${}^n{C}_0={}^n{C}_n=1$$

• If n = 2k, the middle coefficient is

$${=}^n{C}_k=\frac{(2k)!}{(k!{)}^2\ast 2^k}$$

##3. Summations/Identities

$$\sum _{i=0}^n{}^n{C}_i=2^n$$

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

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

• 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):

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

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

6. Last term

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

7. General Term:

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

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, $$T_{\frac{n}{2}+1}={}^n{C}_{\frac{n}{2}-1}{x}^{n-\frac{n}{2}+1}{\left(\frac{a}{x}\right)}^{\frac{n}{2}-1}$$

$$Rightarrow12870={}^n{C}_{\frac{n}{2}-1}{x}^{\frac{n}{2}+1}{a}^{\frac{n}{2}-1}$$

$$Rightarrow12870=\frac{n!}{(\frac{n}{2}-1)!(\frac{n}{2}+1)!}{x}^2{a}^{\frac{n}{2}}-1$$

Here we can substitute (n = 12) to satisfy the given 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:

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

Given:

• (T_1 = a = {}^nC_0a^n)

• (T_3 = b = {}^nC_2a^{n-2}b^2)

• (T_6 = c = {}^nC_5a^{n-5}b^5)

• (T_9=?)

• From $$T_3={}^n{C}_2{a}^{n-2}{b}^2=b$$

$$Rightarrowb={}^n{C}_2{a}^{n-2}{b}^2$$

$$\Rightarrow \frac{b^2}{a^{n-2}}={}^n{C}_2=\frac{n(n-1)}{2}$$

• Similarly from $$T_6={}^n{C}_5{a}^{n-5}{b}^5=c$$

We get,

$$\Rightarrow \frac{c^5}{a^{n-5}{b}^5}={}^n{C}_5=\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}$$

Dividing Equation 2 by Equation 1:

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

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

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

Again from Equation 1,

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

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

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

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

Substituting the value of (a) in Equation (1), we get (9)th term,

$$T_9={}^n{C}_8{a}^{n-8}{b}^8$$ $$={}^n{C}_8{\left(\sqrt[3]{\frac{2b^3}{n(n-1)}}\right)}^{n-8}{b}^8$$

Substituting the value of ({}^nC_8) from the property

$${}^n{C}_8={}^n{C}_{n-8}=\frac{n!}{(n-8)!8!}=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{8!}$$

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

Simplifying, we get

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

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):

$$(3x^2-\frac{1}{2x}{)}^n=\sum _{r=0}^n{}^n{C}_r(3x^2{)}^{n-r}(-\frac{1}{2x}{)}^r$$

$$=\sum _{r=0}^n{}^n{C}_r{3}^{n-r}(-1{)}^r{x}^{2n-2r-r}{2}^{-r}$$

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