JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
If $\sum_{r=1}^{9}\left(\frac{r+3}{2^{r}}\right)\cdot{}^{9}C_r=\alpha\left(\frac{3}{2}\right)^{9}-\beta$, $\alpha,\beta\in\mathbb{N}$, then $(\alpha+\beta)^{2}$ is equal to
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. Split the sum into $\sum r\,{}^{9}C_r\,2^{-r}$ and $3\sum{}^{9}C_r\,2^{-r}$. Formulas used. (1) $r\,{}^{n}C_r=n\,{}^{n-1}C_{r-1}$(2) $\sum_{r=0}^{n}{}^{n}C_r\,x^r=(1+x)^n$Step 1 - First part. By (1), $$\sum_{r=1}^{9}r\,{}^{9}C_r\,2^{-r}=\frac92\sum_{r=1}^{9}{}^{8}C_{r-1}\,2^{-(r-1)}=\frac92\left(\frac32\right)^8=3\left(\frac32\right)^9.$$ Step 2 - Second part. By (2) with $x=\frac12$ (the $r=0$ term is missing): $$3\sum_{r=1}^{9}{}^{9}C_r\,2^{-r}=3\left[\left(\frac32\right)^9-1\right].$$ Step 3 - Add. $$\text{sum}=6\left(\frac32\right)^9-3,$$ so $\alpha=6$, $\beta=3$. Step 4 - Value. $(\alpha+\beta)^2=81$. Answer: (D)