Let $\alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$ and $\beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$. If $140

Question

Let $\alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$ and $\beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}$. If $140 < \frac{2 \alpha}{\beta} < 281$, then the value of $n$ is _______

Aicharya · Accepted Answer

$\begin{aligned} & \alpha=\sum_{r=0}^n\left(4 r^2+2 r+1ight) \cdot{ }^n C_r \ & \alpha=4 \sum_{r=0}^n r^2 \cdot \frac{n}{r} \cdot{ }^{n-1} C_{r-1}+2 \sum_{r=0}^n r \cdot \frac{n}{r} \cdot{ }^{n-1} C_{r-}+\sum_{r=0}^n{ }^n C_r \ & +4 n \sum_{r=0}^n{ }^{n-1} C_{r-1}+2 n \sum_{r=0}^n{ }^{n-1} C_{r-1}+\sum_{r=0}^n{ }^n C_r \ & \alpha=4 n(n-1) \cdot 2^{n-2}+4 n \cdot 2^{n-1}+2 n \cdot 2^{n-1}+2^n \ & \alpha=2^{n-2}[4 n(n-1)+8 n+4 n+4] \ & \alpha=2^{n-2}\left[4 n^2+8 n+4ight] \ & \alpha=2 n(n+1)^2 \ & \beta=\sum_{r=0}^n \frac{{ }^n C_r}{r+1}+\frac{1}{n+1} \ & =\sum_{r=0}^n \frac{C^{n+1} C_{r+1}}{n+1}+\frac{1}{n+1} \ & =\frac{1}{n+1}\left(1+{ }^{n+1} C_1+\ldots .+{ }^{n+1} C_{n+1}ight) \ & =\frac{2^{n+1}}{n+1} \ & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n+1)^2}{2^{n+1}} \cdot(n+1)=(n+1)^3 \ & 140 < (n+1)^3 < 281 \ & n=4 \Rightarrow(n+1)^3=125 \ & n=5 \Rightarrow(n+1)^3=216 \ & n=6 \Rightarrow(n+1)^3=343 \ & 	herefore n=5\end{aligned}$

Let $\alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r$ and $\beta=\left(\sum_{r=0}^n \frac{{ }^n…

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