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.