Question

# Find the sum of $$n$$ terms of the series whose $$n^{th}$$ term is $$n^{3} + \dfrac {3}{2}n$$.

Solution

## $$T_n = n^3 + \dfrac{3}{2}n$$Let $$S_n$$ denotes the sum of the $$n$$ terms of the given series, then$$S_n = \sum\limits_{k=1}^n T_k = \sum\limits_{k=1}^n \left( k^3 + \dfrac{3}{2}k \right)$$$$= \left( \sum\limits_{k=1}^n k^3 \right) + \dfrac{3}{2} \left( \sum\limits_{k=1}^n k \right)$$$$= \left[ \dfrac{n(n+1)}{2} \right] ^2 + \dfrac{3}{2} \left[ \dfrac{n(n+1)}{2} \right]$$$$= \left[ \dfrac{n(n+1)}{2} \right] \left[ \dfrac{n(n+1)}{2} + \dfrac{3}{2} \right]$$$$= \dfrac{n(n+1)(n^2+n+3)}{4}$$Mathematics

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