Question

# If $$S_n=\displaystyle \sum_{r=1}^{n}\dfrac{2r+1}{r^4+2r^3+r^2}$$, then $$S_{10}$$ is equal to

A
1
B
121120
C
120121
D
6061

Solution

## The correct option is C $$\dfrac{120}{121}$$$$\displaystyle S_{n}= \sum_{r=1}^{n}\frac{2r+1}{r^{4}+2r^{3}+r^{2}}=\sum_{r=1}^{n}\frac{2r+1}{r^{2}(r^{2}+2r+1)}= \sum_{r=1}^{n}\frac{2r+1}{r^{2}(r+1)^{2}}$$$$\displaystyle \Rightarrow S_{n}=\sum_{r=1}^{n}[\frac{1}{r^{2}}-\frac{1}{(r+1)^{2}}]$$$$\displaystyle S_{10}= [\frac{1}{1^{2}}-\frac{1}{2^{2}})+[\frac{1}{2^{2}}-\frac{1}{3^{2}})+...+(\frac{1}{9^{2}}-\frac{1}{10^{2}})+(\frac{1}{10^{2}}-\frac{1}{11^{2}})$$$$\displaystyle =1-\frac{1}{121}=\frac{120}{121}$$$$\therefore$$ Option C is correctMathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More