Question

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

• A. ${\mathrm{m}}^2-\mathrm{m}(4\mathrm{r}-1)+4{\mathrm{r}}^2+2=0$
• B. ${\mathrm{m}}^2-\mathrm{m}(4\mathrm{r}-1)+4{\mathrm{r}}^2-2=0$
• C. ${\mathrm{m}}^2-\mathrm{m}(4\mathrm{r}+1)+4{\mathrm{r}}^2-2=0$
• D. ${\mathrm{m}}^2-n(4\mathrm{r}-1)+4{\mathrm{r}}^2-2=0$

Answer 1

The correct option is C

${\mathrm{m}}^2-\mathrm{m}(4\mathrm{r}+1)+4{\mathrm{r}}^2-2=0$

Explanation for the correct option:

Given that, $r^{\text{th}},(r+1{)}^{\text{th}}$ and $(r+2{)}^{\text{th}}$term in the expansion of ${\left(1+y\right)}^m$ are in A.P.

Since, ${}^{\mathrm{m}}\mathrm{C}_{\mathrm{r}-1},{}^{\mathrm{m}}\mathrm{C}_{\mathrm{r}},{}^{\mathrm{m}}\mathrm{C}_{\mathrm{r}+1}$ are in A.P.
$\begin{array}{rcl}{2}^m{C}_r& =& {}^{m}C_{r-1}+{}^{m}C_{r+1}\\ 2& =& \frac{{}^{m}C_{r-1}}{{}^{m}C_r}+\frac{{}^{m}C_{r+1}}{{}^{m}C_r}\\ 2& =& \frac{{\displaystyle \frac{m!}{\left(r-1\right)!\left(m-r+1\right)!}}}{{\displaystyle \frac{m!}{r!\left(m-r\right)!}}}+\frac{{\displaystyle \frac{m!}{\left(r+1\right)!\left(m-r-1\right)!}}}{{\displaystyle \frac{m!}{r!\left(m-r\right)!}}}\\ 2& =& \frac{r}{m-r+1}+\frac{m-r}{r+1}\left(\because \frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{n-r+1}{r}\right)\\ 2& =& \frac{r\left(r+1\right)+\left(m-r\right)\left(m-r+1\right)}{\left(m-r+1\right)\left(r+1\right)}\\ 2& =& \frac{r^2+r+m^2-mr+m-mr+r^2-r}{mr+m-r^2-r+r+1}\\ 2& =& \frac{2r^2+m^2-2mr+m}{mr-r^2+m+1}\\ m^2-m(4r+1)+4r^2-2& =& 0\end{array}$
Hence, option (C) is correct.