Q. If the coefficient 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
Explanation
(b)
Since, the coefficient of given terms are $\ ^{m}C_{r - 1},\ \ ^{m}C_{r},\ \ ^{m}C_{r + 1}$ respectively and they are in AP.
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{m}C_{r - 1} + \ \ ^{m}C_{r + 1} = 2\ ^{m}C_{r}$
$\Longrightarrow \frac{m!}{(r - 1)!(m - r + 1)!} + \frac{m!}{(r + 1)!(m - r - 1)!} = 2\frac{m!}{r!(m - r)!}\$
$\Longrightarrow \frac{1}{(m - r + 1)(m - r)} + \frac{1}{(r + 1)r} = \frac{2}{r!(m - r)}$
$\Longrightarrow \frac{r(r + 1) + (m - r + 1)(m - r)}{r(r + 1)(m - r + 1)(m - r)} = \frac{2}{r(m - r)}$
$\Longrightarrow r^{2} + r + m^{2} + r^{2} - 2mr + m - r$
$= 2\left( mr - r^{2} + r + m - r + 1 \right)$
$\Longrightarrow {4r}^{2} - 4mr - m - 2 + m^{2} = 0$
$\Longrightarrow m^{2} - m(4r + 1) + {4r}^{2} - 2 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.