Question

If three consecutive coefficient in the expansion of $(1+x^n)$ are in the ratio 6 : 33 : 110, Find n.

Answer 1

using given condition

$n_{C_{\gamma }}{:}^n{c}_{{{\rm Y}}_{+1}}{:}^n{C}_{\gamma +2}=$ $6:33:110$

$\frac{n!}{r!(n-2)!}=6x$

$\frac{n!}{(r+1)!(n-r-1)!}=33x$

$\frac{n!}{(r+2)!(n-r-2)!}=110x$

$\frac{(x+1)!(n-x-1)!}{x!(n-8)!}=\frac{6}{33}$

$\Rightarrow 33r+33=6n-6r$

$39r-6n=-33-\left(i\right)$

similarly

$\frac{(r+2)}{(n-r-1)}=\frac{33}{110}$

$\Rightarrow 143r-33n=-253-\left(ii\right)$

Solving (i) and (ii)

$n=12$