Question

If $S_n=\sum _{r=1}^n\frac{2r+1}{r^4+2r^3+r^2}$, then $S_{10}$ is equal to

• A. $1$
• B. $\frac{121}{120}$
• C. $\frac{120}{121}$
• D. $\frac{60}{61}$

Answer 1

The correct option is C $\frac{120}{121}$

$S_n=\sum _{r=1}^n\frac{2r+1}{r^4+2r^3+r^2}=\sum _{r=1}^n\frac{2r+1}{r^2(r^2+2r+1)}=\sum _{r=1}^n\frac{2r+1}{r^2(r+1{)}^2}$

$\Rightarrow S_n=\sum _{r=1}^n[\frac{1}{r^2}-\frac{1}{(r+1{)}^2}]$

$S_{10}=[\frac{1}{1^2}-\frac{1}{2^2})+[\frac{1}{2^2}-\frac{1}{3^2})+...+(\frac{1}{9^2}-\frac{1}{{10}^2})+(\frac{1}{{10}^2}-\frac{1}{{11}^2})$

$=1-\frac{1}{121}=\frac{120}{121}$

$\therefore$ Option C is correct