The sum of the series 1 + 2 times 3 + 3 times | vedclass

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(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \s

Vedclass · Accepted Answer

$946$

Vedclass · Answer

(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k$.Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:$S_n = 2 \left[ \frac{n(n+1)(2n+1)}{6} ight] - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} - \frac{n(n+1)}{2}$.For $n = 11$:$S_{11} = \frac{11(12)(23)}{3} - \frac{11(12)}{2}$.$S_{11} = 11 	imes 4 	imes 23 - 11 	imes 6$.$S_{11} = 1012 - 66 = 946$.

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is:

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