Question

If the coefficient of $r^{th},(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms in the binomial expansion of ${(1+y)}^n$ are in A.P, then $n^2-n(4r+1)+{4r}^2-2$ is equal to

• A. $2$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

Since the coefficient of $r^{th},{(r+1)}^{th},{(r+2)}^{th}$ terms are in A.P., we get
$(\frac{n}{r-1})+(\frac{n}{r+1})=2(\frac{n}{r})$
$\frac{n!}{(n-r+1)!(r-1)!}+\frac{n!}{(n-r-1)!(r+1)!}=2\frac{n!}{(n-r)!r!}$
$\frac{\left[r\right(r+1)+(n-r)(n-r+1)]}{(n-r+1)!(r+1)!}=\frac{2}{(n-r)!r!}$
$r^2+r+n^2-2nr+r^2+n-r=2(n-r+1)(r+1)$
$n^2+2r^2-2nr+n=2(nr+n-r^2-r+r+1)$
$n^2+2r^2-2nr+n=2nr+2n-2r^2+2$
$n^2-n(4r+1)+4r^2-2=0$