Q36 / 60 · KVPY 2011 (SA) · Mathematics
The expression $\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\frac{4^{2}+1}{4^{2}-1}+\ldots \ldots+\frac{(2011)^{2}+1}{(2011)^{2}-1}$ lies in the interval
- A$\left(2010,2010 \frac{1}{2}\right)$
- B$\left(2011-\frac{1}{2011}, 2011-\frac{1}{2012}\right)$
- Correct answerC$\left(2011,2011 \frac{1}{2}\right)$
- D$\left(2012,2012 \frac{1}{2}\right)$
Solution
$\frac{2^{2}+1}{2^{2}-1}+\frac{3^{2}+1}{3^{2}-1}+\frac{4^{2}+1}{4^{2}-1}+\ldots .+\frac{(2011)^{2}+1}{(2011)^{2}-1}$
$\sum_{r=2}^{2011} \frac{r^{2}+1}{r^{2}-1}=\sum_{r=2}^{2011}\left[1+\frac{2}{(r+1)(r-1)}\right]$
$\begin{array}{l}
=\sum_{r=2}^{2011}\left[1+\frac{1}{r-1}-\frac{1}{r+1}\right] \\
=2010+\left[1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\ldots .+\frac{1}{2010}-\frac{1}{2012}\right] \\
=2010+1+\frac{1}{2}-\frac{1}{2012}-\frac{1}{2011} \\
=2011+\frac{1}{2}-\left[\frac{1}{2011}+\frac{1}{2012}\right]
\end{array}$
lies between $\left(2011,2011 \frac{1}{2}\right)$
Hence, the correct answer is option: 'C'.