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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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