lf some three consecutive coefficients in the binomial expansion of (x + 1)ⁿ in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is

Given binomial is (x+ 1)ⁿ, whose general term, is ${\mathrm{T}}_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}$

According to the question, we have

$\begin{array}{r}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}{:}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{:}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}=2:15:70\\ \mathrm{now},\frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}-1}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}=\frac{2}{15}\end{array}$

$\begin{array}{l}\Rightarrow \frac{\frac{\mathrm{n}!}{(\mathrm{r}-1)!(\mathrm{n}-\mathrm{r}+1)!}}{\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}}=\frac{2}{15}\Rightarrow \frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}=\frac{2}{15}\\ \Rightarrow 15\mathrm{r}=2\mathrm{n}-2\mathrm{r}+2\\ \Rightarrow 2\mathrm{n}-17\mathrm{r}+2=0----\left(\mathrm{i}\right)\\ \Rightarrow \text{ similarly, }\frac{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}}=\frac{15}{70}\Rightarrow \frac{\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}}{\frac{\mathrm{n}!}{(\mathrm{r}+1)!(\mathrm{n}-\mathrm{r}-1)!}}=\frac{3}{14}\\ \Rightarrow \frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}}=\frac{3}{14}\Rightarrow 14\mathrm{r}+14=3\mathrm{n}-3\mathrm{r}\\ \Rightarrow 3\mathrm{n}-17\mathrm{r}-14=0----\left(\mathrm{ii}\right)\end{array}$

On solving Eqs. (i) and (ii), we get

$\mathrm{n}-16=0\Rightarrow \mathrm{n}=16\text{ and }\mathrm{r}=2$

Now, the average $=\frac{{ }^{16}{\mathrm{C}}_1{+}^{16}{\mathrm{C}}_2{+}^{16}{\mathrm{C}}_3}{3}$

                           $=\frac{16+120+560}{3}=\frac{696}{3}=232$