Question

If the coefficients of $rth$, $(r+1)$ th and $(r+2)$ th terms in the binomial expansion of $(1+y{)}^m$ are in AP, then $m$ and $r$ satisfy the equation :

• A. $m^2-m(4r-1)+4r^2+2=0$
• B. $m^2-m(4r+1)+4r^2-2=0$
• C. $m^2-m(4r+1)+4r^2+2=0$
• D. $m^2-m(4r-l)+4c^2-2=0$

Answer 1

Answer: (B)

Key Idea : (1) The coefficient of (r + l)th term of $1l+y{)}^m$ is ${}^m{C}_r$
(2) If a, b,c are in AP, then $b=\frac{a+c}{2},$
Sincc, ${}^m{C}_{r-1}{+}^m{C}_{r+1}=2{}^m{C}_r$
$\Rightarrow \frac{m!}{\left(r-1\right)!\left(m-r+1\right)!}+\frac{m!}{\left(r-1\right)!\left(m-r-1\right)!}=2\frac{m!}{r!\left(m-r\right)!}$
$\Rightarrow \frac{1}{\left(m-r+1\right)\left(m-r\right)}+\frac{1}{\left(r+1\right)r}=\frac{2}{r\left(m-r\right)}$
$\Rightarrow \frac{r\left(r+1\right)+\left(m-r+1\right)\left(m-r\right)}{r\left(r+1\right)\left(m-r+1\right)\left(m-r\right)}=\frac{2}{r\left(m-r\right)}$
$\Rightarrow r^2+r+m^2+r^2-2mr+m-r$$=2\left(mr-r^2+r+m-r+1\right)$
$\Rightarrow 4r^2-4mr-m-2+m^2=0$
$\Rightarrow m^2-m\left(4r+1\right)+4r^2-2=0$