Let alpha = sum_{r=0}^{n} (4r^2+2r+1) {}^{n}C | vedclass

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(C) We have $\alpha = \sum_{r=0}^{n} (4r^2+2r+1) {}^{n}C_{r}$.Using the identity $r {}^{n}C_{r} = n {}^{n-1}C_{r-1}$ and $r(r-1) {}^{n}C_{r} = n(n-1)

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$5$

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(C) We have $\alpha = \sum_{r=0}^{n} (4r^2+2r+1) {}^{n}C_{r}$.Using the identity $r {}^{n}C_{r} = n {}^{n-1}C_{r-1}$ and $r(r-1) {}^{n}C_{r} = n(n-1) {}^{n-2}C_{r-2}$,we write $4r^2+2r+1 = 4r(r-1) + 6r + 1$.Then $\alpha = 4n(n-1) \sum_{r=2}^{n} {}^{n-2}C_{r-2} + 6n \sum_{r=1}^{n} {}^{n-1}C_{r-1} + \sum_{r=0}^{n} {}^{n}C_{r}$.$\alpha = 4n(n-1) 2^{n-2} + 6n 2^{n-1} + 2^n = n(n-1) 2^n + 3n 2^n + 2^n = 2^n (n^2 - n + 3n + 1) = 2^n (n^2 + 2n + 1) = 2^n (n+1)^2$.Now,$\beta = \sum_{r=0}^{n} \frac{{}^{n}C_{r}}{r+1} + \frac{1}{n+1} = \sum_{r=0}^{n} \frac{{}^{n+1}C_{r+1}}{n+1} + \frac{1}{n+1} = \frac{1}{n+1} (\sum_{k=1}^{n+1} {}^{n+1}C_{k} + 1) = \frac{1}{n+1} (2^{n+1} - 1 + 1) = \frac{2^{n+1}}{n+1}$.Then $\frac{2\alpha}{\beta} = \frac{2 \cdot 2^n (n+1)^2}{2^{n+1} / (n+1)} = \frac{2^{n+1} (n+1)^2}{2^{n+1} / (n+1)} = (n+1)^3$.Given $140 For $n=4$,$(4+1)^3 = 125$ (too small).For $n=5$,$(5+1)^3 = 216$ (within range).For $n=6$,$(6+1)^3 = 343$ (too large).Thus,$n=5$.

Let $\alpha = \sum_{r=0}^{n} (4r^2+2r+1) {}^{n}C_{r}$ 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...............

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