The sum to 10 terms of the series 1 times 3^{ | vedclass

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(D) 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 $10$ ter

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$13,695$

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(D) 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 $10$ terms is $S_{10} = \sum_{n=1}^{10} (4n^3 + 4n^2 + n)$.Using the standard summation formulas:$\sum_{n=1}^{10} n^3 = \left(\frac{10 	imes 11}{2}ight)^2 = 55^2 = 3025$.$\sum_{n=1}^{10} n^2 = \frac{10 	imes 11 	imes 21}{6} = 385$.$\sum_{n=1}^{10} n = \frac{10 	imes 11}{2} = 55$.Therefore,$S_{10} = 4(3025) + 4(385) + 55$.$S_{10} = 12100 + 1540 + 55 = 13695$.

The sum to $10$ terms of the series $1 \times 3^{2} + 2 \times 5^{2} + 3 \times 7^{2} + \dots$ is

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