Topic
Consider the expansion \({\left( {\frac{x}{2} + \frac{1}{5}} \right)^{21}}\)
a) Determine the coefficient of \({x^{10}}\)
b) State the general term in the above binomial expansion, then state the coefficient \({a_k}\) of \({x^k}\) with \(0 \le k \le 21\)
Solution method – See details
Newton’s binomial formula: \({(a + b)^n} = C_n^0{a^n} + C_n^1{a^{n – 1}}b + … + C_n^{n – 1} a{b^{n – 1}} + C_n^n{b^n}\)
Detailed explanation
a) According to Newton’s binomial formula, we have:
\({\left( {\frac{x}{2} + \frac{1}{5}} \right)^{21}} = C_{21}^0{\left( {\frac{x} {2}} \right)^{12}} + C_{21}^1{\left( {\frac{x}{2}} \right)^{20}}{\left( {\frac{1 }{5}} \right)^1} + … + C_{21}^k{\left( {\frac{x}{2}} \right)^{21 – k}}{\left( {\ frac{1}{5}} \right)^k} + … + C_{21}^{21}{\left( {\frac{1}{5}} \right)^{21}}\)
The term containing \({x^{10}}\) corresponds to \(21 – k = 10 \Rightarrow k = 11\). Therefore the coefficient of \({x^{10}}\) is
\(C_{21}^{11}{\left( {\frac{1}{2}} \right)^{10}}{\left( {\frac{1}{5}} \right)^ {11}}\)
b) Term containing \({x^k}\) in the expansion of \({\left( {\frac{x}{2} + \frac{1}{5}} \right)^{21} }\) is \(C_{21}^{21 – k}{\left( {\frac{x}{2}} \right)^k}{\left( {\frac{1}{5}} \right)^{21 – k}}\)
Thus, the coefficient \({a_k}\) of \({x^k}\) with \(0 \le k \le 21\) is \(C_{21}^{21 – k}{\left ( {\frac{1}{2}} \right)^k}{\left( {\frac{1}{5}} \right)^{21 – k}}\)
