JEE Main 2024 — Sequences And Series Question with Solution

$$\begin{array}{rl}\alpha =& \sum _{r=0}^n\left(4r^2+2r+1\right){}^n{C}_r\\ & =4\sum _{r=0}^n{r}^2{}^n{C}_r+2\sum _{r=0}^{n}r\cdot {}^n{C}_r+\sum _{r=0}^n{}^n{C}_r\\ & =4n(n+1)2^{n-2}+2\cdot n\cdot 2^{n-1}+2^n\\ & =2^n(n(n+1)+n+1)=2^n(n+1{)}^2\\ & \beta =\sum _{r=0}^n\left(\frac{{}^n{C}_r}{r+1}\right)+\left(\frac{1}{n+1}\right)\\ & (1+x{)}^n=\sum _{r=0}^n{}^n{C}_r{x}^r\end{array}$$

\begin{aligned} & \int_\limits0^1(1+x)^n d x=\left.\sum_{r=0}^n \frac{{ }^n C_r x^{r+1}}{r+1}\right|_0 ^1=\sum_\limits{r=0} \frac{{ }^n C}{r+1} \\ & \left.\frac{(1+x)^{+1}}{n+1}\right|_0 ^1=\frac{2^n-1}{n+1} \\ \Rightarrow & \beta=\frac{2^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\ \Rightarrow & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n \quad 1)}{\left(\frac{2^{n+1}}{n+1}\right)}=(n+1)^3 \in(140,281) \\ \Rightarrow & (n+1)^3=216 \\ \Rightarrow & n+1=6 \Rightarrow n=5 \end{aligned}