The sum of n terms of the series 1^3+3^3+5^3+ | vedclass

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Question

(A) The $n$-th term of the series is $T_n = (2n-1)^3$.We need to find the sum $S_n = \sum_{k=1}^{n} (2k-1)^3$.Expanding the term: $(2k-1)^3 = 8k^3 - 1

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$n^2(2n^2-1)$

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(A) The $n$-th term of the series is $T_n = (2n-1)^3$.We need to find the sum $S_n = \sum_{k=1}^{n} (2k-1)^3$.Expanding the term: $(2k-1)^3 = 8k^3 - 12k^2 + 6k - 1$.Now, $S_n = \sum_{k=1}^{n} (8k^3 - 12k^2 + 6k - 1) = 8 \sum k^3 - 12 \sum k^2 + 6 \sum k - \sum 1$.Using the standard summation formulas:$\sum k^3 = \frac{n^2(n+1)^2}{4}$, $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$, $\sum k = \frac{n(n+1)}{2}$, $\sum 1 = n$.Substituting these values:$S_n = 8 \left[ \frac{n^2(n+1)^2}{4} \right] - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.$S_n = 2n^2(n^2+2n+1) - 2n(2n^2+3n+1) + 3n^2+3n - n$.$S_n = 2n^4 + 4n^3 + 2n^2 - 4n^3 - 6n^2 - 2n + 3n^2 + 2n$.$S_n = 2n^4 - n^2 = n^2(2n^2-1)$.

The sum of $n$ terms of the series $1^3+3^3+5^3+7^3+\ldots$ is

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