If sum_{r=0}^5 frac{{}^{11}C_{2r+1}}{2r+2} = | vedclass

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(B) Consider the expansion $(1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k$. Integrating both sides from $0$ to $1$ gives $\int_0^1 (1+x)^{11} dx = \sum_

Vedclass · Accepted Answer

$2035$

Vedclass · Answer

(B) Consider the expansion $(1+x)^{11} = \sum_{k=0}^{11} {}^{11}C_k x^k$. Integrating both sides from $0$ to $1$ gives $\int_0^1 (1+x)^{11} dx = \sum_{k=0}^{11} \frac{{}^{11}C_k}{k+1} = \frac{2^{12}-1}{12}$.Integrating from $-1$ to $0$ gives $\int_{-1}^0 (1+x)^{11} dx = \sum_{k=0}^{11} \frac{{}^{11}C_k (-1)^k}{k+1} = \frac{1}{12}$.Subtracting the two results: $\sum_{k=0}^{11} \frac{{}^{11}C_k (1 - (-1)^k)}{k+1} = \frac{2^{12}-1-1}{12} = \frac{2^{12}-2}{12} = \frac{2^{11}-1}{6}$.The sum $\sum_{k=0}^{11} \frac{{}^{11}C_k (1 - (-1)^k)}{k+1}$ only includes terms where $k$ is odd,i.e.,$k = 2r+1$.Thus,$2 \sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2^{11}-1}{6}$,which implies $\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2^{11}-1}{12} = \frac{2047}{12}$.Here $m = 2047$ and $n = 12$. Since $	ext{gcd}(2047, 12) = 1$,$m - n = 2047 - 12 = 2035$.

If $\sum_{r=0}^5 \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{m}{n}$,$\text{gcd}(m, n) = 1$,then $m - n$ is equal to . . . . . .

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