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A sequence has nth term given by T n =(n 2 +1)n! . Find summation (sigma) T n .

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one month ago

```							Since according to your question the Nth term is i.e. Tn= (n^2+1)n.Apply the special series and sequence formula you will get your answer. Tn = (n^2+1)n = n^3+n Sn= sigma(k=1 to k = n) k^3+ sigma(k=1 to k=n) k Sn= [ n(n+1)/2)]^2 + n(n+1)/2Sn= n(n+1)/2 * [n(n+1)/2 + 2]Sn= n(n+1)(n^2+n+2)/4Hence thats your answer. Thank you very much.
```
one month ago
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