Question

If the coefficients of $r^{th}$, $(r+1{)}^{th}$ and $(r+2{)}^{th}$ terms in the binomial expansion of $(1+y{)}^m$are in $A.P.$, then $m$ and $r$ will 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

The correct option is D $m^2-m(4r+1)+4r^2-2=0$

Here, the coefficients of $T_r$, $T_{r+1}$ and $T_{r+2}$ of $(1+y{)}^m$ are in $A.P.$

$\Rightarrow {}^m{C}_{r-1},{}^m{C}_r$ and ${}^m{C}_{r+1}$ are in $A.P.$

$\Rightarrow 2{}^m{C}_r={}^m{C}_{r-1}+{}^m{C}_{r+1}\Rightarrow 2\frac{m!}{r!(m-r)!}=\frac{m!}{(r-1)!(m-r+1)!}+\frac{m!}{(r+1)!(m-r-1)!}\Rightarrow \frac{2}{r(m-r)}=\frac{1}{(m-r+1)(m-r)}+\frac{1}{(r+1)r}\therefore m^2-m(4r+1)+4r^2-2=0$