Question

If the coefficients of three consecutive terms in the expansion of $(1+x{)}^n$ be 76, 95 and 76, find n.

Answer 1

Suppose r, (r+1) and (r+2) are three consecutive terms in the given expansion.

The coeffcients of these terms are ${}^n{C}_{r-1},{}^n{C}_r$ and ${}^n{C}_{r+1}$

According to the question,

${}^n{C}_{r-1}=76$

${}^n{C}_r=95$

${}^n{C}_{r+2}=76$

$\Rightarrow {}^n{C}_{r-1}={}^n{C}_{r+1}$

$\Rightarrow r-1+r+1=n[I{f}^n{C}_r{=}^n{C}_s\Rightarrow r=sorr+s=n]$

$\Rightarrow r=\frac{n}{2}$

$\therefore \frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{95}{76}$

$\Rightarrow \frac{n-r+1}{r}=\frac{95}{76}\Rightarrow \frac{\frac{n}{2}+1}{\frac{n}{2}}=\frac{95}{76}\Rightarrow 98n+76=\frac{95r}{2}$

$\Rightarrow \frac{19n}{2}=76\Rightarrow n=8$