The sum of the series 1 cdot 3^2 + 2 cdot 5^2 | vedclass

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(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.Expanding this, we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.The sum of the f

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$188090$

Vedclass · Answer

(A) The $n^{th}$ term of the series is given by $T_n = n(2n + 1)^2$.Expanding this, we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.The sum of the first $20$ terms is $S_{20} = \sum_{n=1}^{20} (4n^3 + 4n^2 + n) = 4 \sum_{n=1}^{20} n^3 + 4 \sum_{n=1}^{20} n^2 + \sum_{n=1}^{20} n$.Using the standard summation formulas:$\sum_{n=1}^{N} n^3 = \left[ \frac{N(N+1)}{2} \right]^2$, $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$, and $\sum_{n=1}^{N} n = \frac{N(N+1)}{2}$.For $N = 20$:$4 \sum_{n=1}^{20} n^3 = 4 \left[ \frac{20 \cdot 21}{2} \right]^2 = 4 \cdot (210)^2 = 4 \cdot 44100 = 176400$.$4 \sum_{n=1}^{20} n^2 = 4 \cdot \frac{20 \cdot 21 \cdot 41}{6} = 4 \cdot 2870 = 11480$.$\sum_{n=1}^{20} n = \frac{20 \cdot 21}{2} = 210$.Adding these values: $S_{20} = 176400 + 11480 + 210 = 188090$.

The sum of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ up to $20$ terms is

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