Mathematics - Binomial Theorem Question with Solution | TestHub

We know that.

Binomial expansion of ${\left(2+x\right)}^9$ is given by,

${\left(2+x\right)}^9={}^{9}C_0{x}^0{2}^9+{}^{9}C_1{x}^1{2}^8+...........{}^{9}C_9{x}^9{2}^0$

Now put $x=1$ we get,

${\left(2+1\right)}^9={}^{9}C_0{2}^9+{}^{9}C_1{2}^8++{}^{9}C_2{2}^7+...........{}^{9}C_9{2}^0$

$\Rightarrow 3^9={}^{9}C_0{2}^9+{}^{9}C_1{2}^8++{}^{9}C_2{2}^7+...........{}^{9}C_9{2}^0$

So, sum of coefficient of $x, x^2, x^3, ........., x^7$ will be,

$\Rightarrow {}^{9}C_1{2}^8+{}^{9}C_2{2}^7+...........{}^{9}C_7{2}^2=3^9-{}^{9}C_0{2}^9-{}^{9}C_8{2}^1-{}^{9}C_9{2}^0$

$\Rightarrow {}^{9}C_1{2}^8+{}^{9}C_2{2}^7+...........{}^{9}C_7{2}^2=3^9-2^9-18-1$

$\Rightarrow {}^{9}C_1{2}^8+{}^{9}C_2{2}^7+...........{}^{9}C_7{2}^2=19683-512-19$

$\Rightarrow {}^{9}C_1{2}^8+{}^{9}C_2{2}^7+...........{}^{9}C_7{2}^2=19152$

Now mean is given by $\frac{19152}{7}=2736$