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Vanya Saxena Grade: Upto college level
```        Let sn=∑k=1 to n[n/(n2+kn+k2)] and Tn=∑k=0 to (n-1)  [n/(n2+nk+k2)] for n=1,2,3....... Then
A  Sn‹ pi/3√3                          C  Tn
B  Sn> pi/3√3                           D  Tn>pi/3√3```
8 years ago

147 Points
```										Hi vanya
You can find Sn by summation of series using definite intigral as the limit of sum
sn=∑k=1 to n[n/(n2+kn+k2)]
sn=1/n∑k=1 to n [n2/(n2+kn+k2)]
sn=1/n∑k=1 to n [1/(1+k/n+(k/n)2)]
sn =∫1/(1+x+x2)dx   limit 0 to 1
after solving
sn =pi/3√3

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best vanya   Regards, Askiitians Experts BAdiuddin

```
8 years ago
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