JEE Main · 2005 · 08 May 2005 · Offline · Binomial Theorem

The coefficients of the $r$th, $(r+1)$th and $(r+2)$th terms of $(1+y{)}^m$ are $\left(\genfrac{}{}{0ex}{}{m}{r-1}\right)$, $\left(\genfrac{}{}{0ex}{}{m}{r}\right)$ and $\left(\genfrac{}{}{0ex}{}{m}{r+1}\right)$. For these to be in A.P. we need $2\left(\genfrac{}{}{0ex}{}{m}{r}\right)=\left(\genfrac{}{}{0ex}{}{m}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{m}{r+1}\right)$. Dividing by $\left(\genfrac{}{}{0ex}{}{m}{r}\right)$ and using $\frac{\left(\genfrac{}{}{0ex}{}{m}{r-1}\right)}{\left(\genfrac{}{}{0ex}{}{m}{r}\right)}=\frac{r}{m-r+1}$ and $\frac{\left(\genfrac{}{}{0ex}{}{m}{r+1}\right)}{\left(\genfrac{}{}{0ex}{}{m}{r}\right)}=\frac{m-r}{r+1}$, this becomes $2=\frac{r}{m-r+1}+\frac{m-r}{r+1}$. Clearing the denominators and simplifying gives $m^2-m(4r+1)+4r^2-2=0$.