$2^{3} + 4^{3} + 6^{3} + 8^{3} + . . . .$

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Solution

$2^{3} + 4^{3} + 6^{3} + 8^{3} + . . . . L e t T_{n} b e t h e n t h t e r m o f t h i s s e r i e s . T h e n, T_{n} = (2 n)^{3} = 8 n^{3} L e t S_{n} be the sum to n terms of the given series; Then, S_{n} = \sum_{k - 1}^{n} 8 k^{3} = 8 \sum_{k - 1}^{n} k^{3} = 8 {[\frac{n (n + 1)}{2}]}^{2} = 8 \times \frac{n^{2} (n + 1)^{2}}{4} = 2 n^{2} (n + 1)^{2} H e n c e, S_{n} = 2 n^{2} (n + 1)^{2}$

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