Question

If the coefficient of $r^{th},(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms in the binominal 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. $m^2-m(4r-1)+4r^2-2=0$

Answer 1

Answer: (B)

$m^2-m(4r+1)+4r^2-2=0$

Binomial Expansion of $(1+y{)}^m=1+my+\frac{m(m-1)}{2!}{y}^2+\frac{m(m-1)(m-2)}{3!}{y}^3+\dots$

The $r^{th}$ coefficient in this expansion is ${}^m{C}_{r-1}$

Now, as $r^{th}$ coefficient, $r+1^{th}$ coefficient and $r+2^{th}$ coefficient are in AP,

$2{}^m{C}_r={}^m{C}_{r-1}+{}^m{C}_{r+1}$

$\therefore 2\frac{m!}{(m-r)!r!}=\frac{m!}{(m-r+1)!(r-1)!}+\frac{m!}{(m-r-1)!(r+1)!}$

$\therefore \frac{2}{(m-r)r}=\frac{1}{(m-r+1)(m-r)}+\frac{1}{r(r+1)}$

$\therefore \frac{1}{(m-r)r}-\frac{1}{r(r+1)}=\frac{1}{(m-r+1)(m-r)}-\frac{1}{(m-r)r}$

$\therefore \frac{r+1-m+r}{(m-r)r(r+1)}=\frac{r-m+r-1}{r(m-r+1)(m-r)}$

$\therefore \frac{2r-m+1}{r+1}=\frac{2r-m-1}{(m-r+1)}$

$\therefore (2r-m+1)(m-r+1)=(2r-m-1)(r+1)$

$\therefore 2mr-2r^2+2r-m^2+mr-m+m-r+1=2r^2+2r-mr-m-r-1$

$\therefore m^2+4r^2-m(4r+1)-2=0$