Question

If some three consecutive in the binomial expansion of ${(x+1)}^n$ is powers of $x$ are in the ratio $2:15:70$, then the average of these three coeffcient is:

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

Answer 1

The correct option is D $232$

Let the terms be ${}^n{C}_{r-1},{}^n{C}_r,{}^n{C}_{r+1}$

$\Rightarrow {}^n{C}_{r-1}:{}^n{C}_r:{}^n{C}_{r+1}=2:15:70$

$\frac{{{}^{n}C}_{r-1}}{{{}^{n}C}_r}=\frac{2}{15}$

$\frac{\frac{n!}{(r-1)(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{2}{15}$

$\frac{r}{n-r+1}=\frac{2}{15}\Rightarrow 17r=2n+2$

$\frac{{{}^{n}C}_r}{{{}^{n}C}_{r+1}}=\frac{15}{70}$

$\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}}=\frac{3}{14}$

$\frac{r+1}{n-r}=\frac{3}{14}$

$14r+14=3n-3r$

$3n-17r=14;2n-17r=-2\Rightarrow n=16$

$17r=34;r=2$

${{}^{16}C}_1,{{}^{16}C}_2,{{}^{16}C}_3$

$\frac{{{}^{16}C}_1+{{}^{16}C}_2+{{}^{16}C}_3}{3}=\frac{16+120+560}{3}=232$