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 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 $^mC_r$
(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)$
$\Rightarrow 4r^{2}-4mr-m-2+m^{2}=0$
$\Rightarrow m^{2} - m \left(4r + 1\right) + 4r^{2} -2=0$