The sum of the series frac{1}{1 times 2} {}^{ | vedclass

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(B) Let $S = \sum_{r=0}^{25} \frac{{}^{25}C_{r}}{(r+1)(r+2)}$.Using the identity $\frac{1}{(r+1)(r+2)} = \int_{0}^{1} x^{r+1} dx$ is not direct,so we

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$\frac{2^{27}-28}{26 	imes 27}$

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(B) Let $S = \sum_{r=0}^{25} \frac{{}^{25}C_{r}}{(r+1)(r+2)}$.Using the identity $\frac{1}{(r+1)(r+2)} = \int_{0}^{1} x^{r+1} dx$ is not direct,so we use $\frac{1}{(r+1)(r+2)} = \int_{0}^{1} \int_{0}^{1} (xy)^{r} x dy dx = \int_{0}^{1} \int_{0}^{1} (xy)^{r} x dy dx$ or simply integrate $(1+x)^{25}$ twice.Consider $(1+x)^{25} = \sum_{r=0}^{25} {}^{25}C_{r} x^{r}$.Integrating from $0$ to $x$: $\int_{0}^{x} (1+t)^{25} dt = \sum_{r=0}^{25} {}^{25}C_{r} \frac{x^{r+1}}{r+1}$.$\frac{(1+x)^{26}-1}{26} = \sum_{r=0}^{25} {}^{25}C_{r} \frac{x^{r+1}}{r+1}$.Now,divide by $x$ and integrate from $0$ to $1$: $\int_{0}^{1} \frac{(1+x)^{26}-1}{26x} dx = \sum_{r=0}^{25} \frac{{}^{25}C_{r}}{r+1} \int_{0}^{1} x^{r} dx = \sum_{r=0}^{25} \frac{{}^{25}C_{r}}{(r+1)(r+2)}$.To evaluate $\frac{1}{26} \int_{0}^{1} \frac{(1+x)^{26}-1}{x} dx$,let $1+x = u$,$dx = du$. When $x=0, u=1$; when $x=1, u=2$.$= \frac{1}{26} \int_{1}^{2} \frac{u^{26}-1}{u-1} du = \frac{1}{26} \int_{1}^{2} (1 + u + u^{2} + \ldots + u^{25}) du$.$= \frac{1}{26} [u + \frac{u^{2}}{2} + \ldots + \frac{u^{26}}{26}]_{1}^{2} = \frac{1}{26} [(2 + \frac{2^{2}}{2} + \ldots + \frac{2^{26}}{26}) - (1 + \frac{1}{2} + \ldots + \frac{1}{26})]$.Alternatively,using the property $\sum_{r=0}^{n} \frac{{}^{n}C_{r}}{(r+1)(r+2)} = \frac{2^{n+2} - (n+2) - 1}{(n+1)(n+2)}$.For $n=25$: $\frac{2^{27} - 27 - 1}{26 	imes 27} = \frac{2^{27} - 28}{26 	imes 27}$.

The sum of the series $\frac{1}{1 \times 2} {}^{25}C_{0} + \frac{1}{2 \times 3} {}^{25}C_{1} + \frac{1}{3 \times 4} {}^{25}C_{2} + \ldots + \frac{1}{26 \times 27} {}^{25}C_{25}$ is

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