Topic
Consider the expansion \({\left( {x + \frac{5}{2}} \right)^{12}}\)
a) Determine the coefficient of \({x^7}\)
b) State the general term in the above binomial expansion, then state the coefficient \({a_k}\) of \({x^k}\) with \(0 \le k \le 12\)
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:
\({\left( {x + \frac{5}{2}} \right)^{12}} = C_{12}^0{x^{12}} + C_{12}^1{x^ {11}}{\left( {\frac{5}{2}} \right)^1} + … + C_{12}^k{x^{12 – k}}{\left( {\frac{ 5}{2}} \right)^k} + … + C_{12}^{12}{\left( {\frac{5}{2}} \right)^{12}}\)
The term containing \({x^7}\) corresponds to \(12 – k = 7 \Rightarrow k = 5\). Therefore the coefficient of \({x^7}\) is
\(C_{12}^5{\left( {\frac{5}{2}} \right)^5}\)
b) The term containing \({x^k}\) in the expansion of \({\left( {x + \frac{5}{2}} \right)^{12}}\) is \(C_) {12}^{12 – k}{(x)^k}{\left( {\frac{5}{2}} \right)^{12 – k}}\)
Thus, the coefficient \({a_k}\) of \({x^k}\) with \(0 \le k \le 12\) is \(C_{12}^{12 – k}{\left ( {\frac{5}{2}} \right)^{12 – k}}\)
