JEE Main 2025 Apr 3 Shift 1, Mathematics Q9: Properties Of Binomial Coefficient

Split the sum:

$S={\sum }_{r=1}^9\frac{r}{2^r}\left(\genfrac{}{}{0ex}{}{9}{r}\right)+3{\sum }_{r=1}^9\frac{1}{2^r}\left(\genfrac{}{}{0ex}{}{9}{r}\right).$

First sum: Using $r\left(\genfrac{}{}{0ex}{}{9}{r}\right)=9\left(\genfrac{}{}{0ex}{}{8}{r-1}\right)$,

${\sum }_{r=1}^9\frac{r}{2^r}\left(\genfrac{}{}{0ex}{}{9}{r}\right)=9{\sum }_{r=1}^9\frac{1}{2^r}\left(\genfrac{}{}{0ex}{}{8}{r-1}\right)=\frac{9}{2}{\sum }_{k=0}^8\frac{1}{2^k}\left(\genfrac{}{}{0ex}{}{8}{k}\right)=\frac{9}{2}{\left(1+\frac{1}{2}\right)}^8=\frac{9}{2}{\left(\frac{3}{2}\right)}^8.$

Second sum:

${\sum }_{r=1}^9\frac{1}{2^r}\left(\genfrac{}{}{0ex}{}{9}{r}\right)={\left(1+\frac{1}{2}\right)}^9-1={\left(\frac{3}{2}\right)}^9-1.$

Combining:

$S=\frac{9}{2}{\left(\frac{3}{2}\right)}^8+3\left[{\left(\frac{3}{2}\right)}^9-1\right]=3{\left(\frac{3}{2}\right)}^9+3{\left(\frac{3}{2}\right)}^9-3=6{\left(\frac{3}{2}\right)}^9-3.$

So $\alpha =6,\beta =3$, and $(\alpha +\beta {)}^2=81.$