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

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(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.Expanding this,we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.The sum of $n$ terms is $S_{n} =

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

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(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.Expanding this,we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.The sum of $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} (8k^{3} - 12k^{2} + 6k - 1)$.Using standard summation formulas:$S_{n} = 8 \left[ \frac{n(n+1)}{2} \right]^{2} - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.$S_{n} = 2n^{2}(n+1)^{2} - 2n(n+1)(2n+1) + 3n(n+1) - n$.$S_{n} = 2n^{2}(n^{2}+2n+1) - 2n(2n^{2}+3n+1) + 3n^{2} + 3n - n$.$S_{n} = 2n^{4} + 4n^{3} + 2n^{2} - 4n^{3} - 6n^{2} - 2n + 3n^{2} + 2n$.$S_{n} = 2n^{4} - n^{2} = n^{2}(2n^{2}-1)$.

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

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