The mean of the coefficients of [x, x^2, , x^7] in the binomial expansion of (2+x)^9 is __…

Question

The mean of the coefficients of $$x, x^{2}, \ldots, x^{7}$$ in the binomial expansion of $(2+x)^{9}$ is \_\_\_\_\_\_\_\_\_\_\_.

Answer: 2736

Accepted Answer

We have, binomial coefficient, $(2+x{)}^{9}$

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

Coefficient of $x({T}_{1})={}^{9}{C}_{1}	imes {2}^{8}$

Coefficient of ${x}^{2}({T}_{2})={}^{9}{C}_{2}	imes {2}^{7}$

Coefficient of ${x}^{3}({T}_{3})={}^{9}{C}_{3}	imes {2}^{6}$

.                    .

Coefficient of ${x}^{7}({T}_{7})={}^{9}{C}_{7}	imes {2}^{2}$

$=\frac{{}^{9}{C}_{1}	imes {2}^{8}+{}^{9}{C}_{2}	imes {2}^{7}+{}^{9}{C}_{3}	imes {2}^{6}+\ldots +{}^{9}{C}_{7}	imes {2}^{2}}{7}$

$\begin{gathered} {}^{9}{C}_{0}	imes {2}^{9}+{}^{9}{C}_{1}	imes {2}^{8}+{}^{9}{C}_{2}	imes {2}^{7}+{}^{9}{C}_{3}	imes {2}^{6}+\ldots .+{}^{9}{C}_{7}	imes {2}^{2} \ =\frac{+{}^{9}{C}_{8}	imes {2}^{1}+{}^{9}{C}_{9}	imes {2}^{0}-{}^{9}{C}_{0}	imes {2}^{9}-{}^{9}{C}_{8}	imes {2}^{1}-{}^{9}{C}_{9}	imes {2}^{0}}{7} \end{gathered}$

$=\frac{(1+2{)}^{9}-{2}^{9}-18-1}{7}=\frac{19152}{7}=2736$

Solutions · 1