Question

If 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 :

• A. $227$
• B. $232$
• C. $625$
• D. $964$

Answer 1

The correct option is B $232$

Let the coefficients of terms $t_r,t_{r+1},t_{r+2}$ are in ratio $2:15:70$.
$\therefore {}^n{C}_{r-1}:{}^n{C}_r:{}^n{C}_{r+1}=2:15:70$

$\Rightarrow \frac{{}^n{C}_{r-1}}{{}^n{C}_r}=\frac{2}{15}\Rightarrow \frac{r}{n-r+1}=\frac{2}{15}$
$\Rightarrow 15r=2n-2r+2$
$\Rightarrow 17r=2n+2\dots \left[1\right]$

Also, $\frac{{}^n{C}_r}{{}^n{C}_{r+1}}=\frac{15}{70}\Rightarrow \frac{r+1}{n-r}=\frac{15}{70}$
$\Rightarrow 14r+14=3n-3r$
$\Rightarrow 17r=3n-14\dots \left[2\right]$

From $\left[1\right]$ and $\left[2\right]$, we get
$n=16$ and $r=2$

First term $={}^{16}{C}_1=16$
Therefore, other two terms are ${}^{16}{C}_2=120$ and ${}^{16}{C}_3=560$.
Therefore, average of coefficients $=232$