The expression frac{2^2+1}{2^2-1}+frac{3^2+1} | vedclass

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(C) Let $S = \sum_{r=2}^{2011} \frac{r^2+1}{r^2-1}$.We can rewrite the general term $T_r$ as:$T_r = \frac{r^2-1+2}{r^2-1} = 1 + \frac{2}{(r-1)(r+1)}$.

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$\left(2011, 2011 \frac{1}{2}\right)$

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(C) Let $S = \sum_{r=2}^{2011} \frac{r^2+1}{r^2-1}$.We can rewrite the general term $T_r$ as:$T_r = \frac{r^2-1+2}{r^2-1} = 1 + \frac{2}{(r-1)(r+1)}$.Using partial fractions:$T_r = 1 + \left(\frac{1}{r-1} - \frac{1}{r+1}\right)$.Summing from $r=2$ to $2011$:$S = \sum_{r=2}^{2011} 1 + \sum_{r=2}^{2011} \left(\frac{1}{r-1} - \frac{1}{r+1}\right)$.The first part is $\sum_{r=2}^{2011} 1 = 2010$.The second part is a telescoping sum:$\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \ldots + \left(\frac{1}{2009} - \frac{1}{2011}\right) + \left(\frac{1}{2010} - \frac{1}{2012}\right)$.After cancellation,we get:$1 + \frac{1}{2} - \frac{1}{2011} - \frac{1}{2012} = \frac{3}{2} - \left(\frac{1}{2011} + \frac{1}{2012}\right)$.Thus,$S = 2010 + 1.5 - \left(\frac{1}{2011} + \frac{1}{2012}\right) = 2011.5 - \left(\frac{1}{2011} + \frac{1}{2012}\right)$.Since $0 < \left(\frac{1}{2011} + \frac{1}{2012}\right) < 1$,the value of $S$ lies in the interval $\left(2011, 2011 \frac{1}{2}\right)$.

The expression $\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\frac{4^2+1}{4^2-1}+\ldots+\frac{(2011)^2+1}{(2011)^2-1}$ lies in the interval

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