Question

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

• A. $(2010,2010\frac{1}{2})$
• B. $(2011-\frac{1}{2011},2011-\frac{1}{2012})$
• C. $(2011,2011\frac{1}{2})$
• D. $(2012,2012\frac{1}{2})$

Answer 1

The correct option is C $(2011,2011\frac{1}{2})$

$\frac{2^2+1}{2^2-1}+\frac{3^2+1}{3^2-1}+\frac{4^2+1}{4^2-1}........\frac{{\left(2011\right)}^2+1}{{\left(2011\right)}^2-1}=\sum _{n=2}^{2011}\frac{n^2+1}{n^2-1}\sum _{n=2}^{2011}\frac{n^2+1}{n^2-1}=\sum _{n=2}^{2011}\frac{n^2+1-1+1}{n^2-1}=\sum _{n=2}^{2011}\frac{n^2-1+2}{n^2-1}=\sum _{n=2}^{2011}1+\frac{2}{(n+1)(n-1)}=\sum _{n=2}^{2011}1+\sum _{n=2}^{2011}\frac{2}{(n+1)(n-1)}=2010+\sum _{n=2}^{2011}\frac{1}{n-1}-\frac{1}{n+1}=2010+(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}........\frac{1}{2011}-\frac{1}{2012})=2010+1+\frac{1}{2}-\frac{1}{2011}-\frac{1}{2012}=2011+\frac{1}{2}-(\frac{1}{2011}+\frac{1}{2012})$

$=2011.5-(\frac{1}{2011}+\frac{1}{2012})$

Clearly $2011.5-(\frac{1}{2011}+\frac{1}{2012})\in (2011,2011\frac{1}{2})$

Hence, option $C$ is correct.