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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
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B
121120
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C
120121
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D
6061
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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 correct

Mathematics

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