Question

If the coefficients of $r^{th},(r+1{)}^{th},$ and $(r+2{)}^{th}$ terms in the binomial expension of $(1+y{)}^m$ are in A.P. then $\text{m}$ and $\text{r}$ satisfy the equation.

• A. $m^2-m(4r-1)+4r^2+2=0$
• B. $m^2-m(4r+1)+4r^2-2=0$
• C. $m^2-m(4r-1)+(4r^2+3)=0$
• D. $m^2-m(4r-1)+(4r^2-4)=0$

Answer 1

Answer: (B)

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

$\text{Cofficient of}$

$r^{\text{th}},(r+1{)}^{\text{th}}\text{and}(r+2{)}^{\text{th}}\text{terms in the}$

$\text{binomial expansion of}(1+y{)}^m\text{are in A.P.}$

$\text{we know that,}$

$T_{r+1}={}^m{c}_r{y}^r$

$T_r={}^m{c}_{r-1}(y{)}^{r-1}$

$and$

$T_{r+2}=m_{C_{r+1}}(y{)}^{r+1}$

According to condition, we get

${}^m{C}_{r-1},{}^m{C}_r$ and ${}^m{C}_{r+1}$ are in AP

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

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

We know that ${}^m{c}_r=\frac{m!}{r!(m-r)!}$

$\Rightarrow \frac{r}{m-r+1}+\frac{m-r}{r+1}=2$

$\Rightarrow r^2+r+(m-r{)}^2+m-r=2(m-r+1\left)\right(r+1)$

$\Rightarrow r^2+r+m^2+r^2-2mr+m-r=2mr+2m-2r^2-2r+2r+2$

$\Rightarrow 4r^2+m^2-4mr-m-2=0$

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