The value of 1^2 + 3^2 + 5^2 + ....... + 25^2 | vedclass

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Question

(A) The given series is $1^2 + 3^2 + 5^2 + ....... + 25^2$.The $n^{th}$ term is $T_n = (2n - 1)^2$. For the last term $25$,we have $2n - 1 = 25$,which

Vedclass · Accepted Answer

$2925$

Vedclass · Answer

(A) The given series is $1^2 + 3^2 + 5^2 + ....... + 25^2$.The $n^{th}$ term is $T_n = (2n - 1)^2$. For the last term $25$,we have $2n - 1 = 25$,which gives $n = 13$.The sum $S_{13} = \sum_{n=1}^{13} (2n - 1)^2 = \sum_{n=1}^{13} (4n^2 - 4n + 1)$.Using the summation formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$:$S_{13} = 4 \sum_{n=1}^{13} n^2 - 4 \sum_{n=1}^{13} n + \sum_{n=1}^{13} 1$$S_{13} = 4 \left[ \frac{13(13+1)(2 	imes 13 + 1)}{6} ight] - 4 \left[ \frac{13(13+1)}{2} ight] + 13$$S_{13} = 4 \left[ \frac{13 	imes 14 	imes 27}{6} ight] - 2(13 	imes 14) + 13$$S_{13} = 4 	imes (13 	imes 7 	imes 9) - 2(182) + 13$$S_{13} = 4 	imes 819 - 364 + 13$$S_{13} = 3276 - 364 + 13 = 2925$.

The value of $1^2 + 3^2 + 5^2 + ....... + 25^2$ is

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