# Binomial Theorem – Problems

## Binomial Theorem – Problems

Key Points:

1. $${{(x+a)}^{n}}={{n}_{{{C}_{0}}}}{{x}^{n}}+{{n}_{{{C}_{1}}}}{{x}^{n-1}}a+…..+{{n}_{{{C}_{n}}}}{{a}^{n}}$$.

2. $${{T}_{r+1}}={{n}_{{{C}_{r}}}}{{x}^{n-r}}.{{a}^{r}}$$.

3. Number of terms in (x + a) ⁿ is (n + 1), where n is a positive integer

Examples 1: Expand $${{\left( \frac{2p}{5}-\frac{3q}{7} \right)}^{6}}$$ by using binomial theorem

Solution: $${{\left( \frac{2p}{5}-\frac{3q}{7} \right)}^{6}}={{6}_{{{C}_{0}}}}{{\left( \frac{2p}{5} \right)}^{6}}+{{6}_{{{C}_{1}}}}{{\left( \frac{2p}{5} \right)}^{5}}\left( \frac{-3q}{7} \right)+{{6}_{{{C}_{2}}}}{{\left( \frac{2p}{5} \right)}^{4}}{{\left( \frac{-3q}{7} \right)}^{2}}+….$$.

$$\sum\limits_{r=0}^{6}{{{(-1)}^{r}}{{6}_{{{C}_{r}}}}{{\left( \frac{2p}{5} \right)}^{6-r}}{{\left( \frac{3q}{7} \right)}^{r}}}$$.

Examples 2: Find the 10th term in $${{\left( \frac{3p}{4}-5q \right)}^{14}}$$.

Solution: $${{\left( \frac{3p}{4}-5q \right)}^{14}}$$.

We know that $${{T}_{r+1}}={{n}_{{{C}_{r}}}}{{x}^{n-r}}.{{a}^{r}}$$.

$${{T}_{9+1}}={{14}_{{{C}_{9}}}}{{\left( \frac{3p}{4} \right)}^{14-9}}{{\left( -5q \right)}^{9}}$$.

$$=-\frac{14\times 13\times 12\times 11\times 10}{1\times 2\times 3\times 4\times 5}\times \frac{{{3}^{5}}{{p}^{5}}}{{{4}^{5}}}\times {{5}^{9}}{{q}^{9}}$$.

$$=-\frac{2002\times {{3}^{5}}\times {{5}^{9}}}{{{4}^{5}}}\times {{p}^{5}}{{q}^{9}}$$.

Example 3: Find the number of terms in the expansion of

(i) $${{\left( \frac{3a}{4}+\frac{b}{2} \right)}^{8}}$$.

Solution: Number of terms in (x + a)ⁿ  is (n + 1), where n is a positive integer.

Hence number of terms in $${{\left( \frac{3a}{4}+\frac{b}{2} \right)}^{8}}$$.

= 8 + 1 = 9

(ii) (3p + 4q)¹⁴

Solution: Number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer

Hence number of terms in (3p + 4q)¹⁴ = 14 + 1 = 15.