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प्रश्न
13 + 33 + 53 + 73 + ...
उत्तर
Let \[T_n\] be the nth term of the given series.
Thus, we have: \[T_n = \left( 2n - 1 \right)^3\]
Now, let \[S_n\] be the sum of n terms of the given series.
Thus, we have:
\[S_n = \sum^n_{k = 1} T_k \]
\[ = \sum^n_{k = 1} \left[ 2k - 1 \right]^3 \]
\[ = \sum^n_{k = 1} \left[ 8 k^3 - 1 - 6k\left( 2k - 1 \right) \right]\]
\[ = \sum^n_{k = 1} \left[ 8 k^3 - 1 - 12 k^2 + 6k \right]\]
\[ = \sum^n_{k = 1} \left[ 8 k^3 - 1 - 12 k^2 + 6k \right]\]
\[ = {8\sum}^n_{k = 1} k^3 - \sum^n_{k = 1} 1 - 12 \sum^n_{k = 1} k^2 + {6\sum}^n_{k = 1} k \]
\[ = \frac{8 n^2 \left( n + 1 \right)^2}{4} - n - \frac{12n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{6 n\left( n + 1 \right)}{2}\]
\[ = 2 n^2 \left( n + 1 \right)^2 - n - 2n\left( n + 1 \right)\left( 2n + 1 \right) + 3n\left( n + 1 \right)\]
\[ = n\left( n + 1 \right)\left[ 2n\left( n + 1 \right) - 2\left( 2n + 1 \right) + 3 \right] - n\]
\[ = n\left( n + 1 \right)\left[ 2 n^2 - 2n + 1 \right] - n\]
\[ = n\left[ 2 n^3 - 2 n^2 + n + 2 n^2 - 2n + 1 - 1 \right]\]
\[ = n\left[ 2 n^3 - n \right]\]
\[ = n^2 \left[ 2 n^2 - 1 \right]\]
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