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Let for $\mathrm{n}=1,2, \ldots . ., 50, \mathrm{~S}_{\mathrm{n}}$ be the sum of the infinite geometric progression whose first term is $\mathrm{n}^{2}$ and whose common ratio is $\mathrm{\frac{1}{(n+1)^{2}}}$ . Then the value of $\frac{1}{26}+\sum_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to____________
Option: 1
41651

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{a =n^{2}, \quad r=\frac{1}{(n+1)^{2}}} \\$

$\mathrm{\therefore S_{n} =\frac{n^{2}}{1-\frac{1}{(n+1)^{2}}}=\frac{n^{2}(n+1)^{2}}{n^{2}+2 n}} \\$

$\mathrm{=\frac{n(n+1)^{2}}{n+2}=\frac{n\left(n^{2}+2 n+1\right)}{n+2}}$

Using long division

$\mathrm{S_{n}=n^{2}+1-\frac{2}{n+2}} \\$

$\mathrm{\therefore S_{n}+\frac{2}{n+1}-n-1} \\$

$\mathrm{= n^{2}+1-\frac{2}{n+2}-n-1+\frac{2}{n+1}} \\$

$\mathrm{= n^{2}-n+2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]}$

Apply summation

$\mathrm{\sum_{1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)} \\$

$\mathrm{=\sum n^{2}-\sum n+2 \sum\left(\frac{1}{n+i}-\frac{1}{n+2}\right) }\\$

$\mathrm{=\frac{50 \times 51 \times 101}{6}-\frac{50 \times 51}{2}-2\left(\frac{1}{52}-\frac{1}{2}\right) }$

$\mathrm{\therefore \frac{1}{26}+\sum\left(S_{n}+\frac{2}{n+1}-n-1\right)} \\$

$\mathrm{=\frac{1}{26}+\frac{50 \times 51\times 101}{6}-\frac{50 \times 51}{2}-2\left(\frac{1}{52}-\frac{1}{2}\right) }\\$

$\mathrm{=41651 }$

Hence answer is 41651

Posted by

Ritika Kankaria

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