Question

If the coefficients of $r^{th},{(r+1)}^{th},{(r+2)}^{th}$ terms in ${(1+x)}^n$ are in A.P. then show that $n^2-(4r+1)n+4r^2-2=0$.

Answer 1

Given $r^{th},(4+1{)}^{th},$ & $(r+2{)}^{th}$ term of expansion.

$(1+x{)}^n$ are in AP then,

$r^{th}term=n_{c_{r-1}}(r+1{)}^{th}tan=n_{c_r}$

$(r+2{)}^{th}tan=n_{c_{r+1}}$

now $n_{c_{r-1}},n_{c_r},n_{c_{r+1}}$ are in AP then,

$\Rightarrow n_{c_r}-n_{c_{r-1}}=n_{c_{r+1}}-n_{c_r}$

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

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

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

$=\frac{1}{n-r}[\frac{1}{r}-\frac{1}{n-r+1}]=\frac{1}{r}[\frac{1}{r+1}-\frac{1}{n-r}]$

$=\frac{n-r+1-r}{r(n-r)(n-r+1)}=\frac{1}{r}\left[\frac{n-r-r-1}{(r+1)(n-r)}\right]$

$\Rightarrow (n-2r+1)(r+1)=(n-2r-1)(n-r+1)$

$\Rightarrow nr+n-2r^2-2r+r+1=n^2-nr+n-2nr+2r^2-2r-n+r-1$

$\Rightarrow n^2-4nr+4r^2-n-2=0$

$\Rightarrow n^2-(4r+1)n+4r^2-2=0$ Hence Prove