The value of {1^2} + {3^2} + {5^2} + dots + { | vedclass

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(A) The given series is the sum of squares of the first $13$ odd numbers.The general term is ${T_n} = {(2n - 1)^2}$ for $n = 1, 2, \dots, 13$.The sum

Vedclass · Accepted Answer

$2925$

Vedclass · Answer

(A) The given series is the sum of squares of the first $13$ odd numbers.The general term is ${T_n} = {(2n - 1)^2}$ for $n = 1, 2, \dots, 13$.The sum is $S = \sum_{n=1}^{13} (2n - 1)^2 = \sum_{n=1}^{13} (4n^2 - 4n + 1)$.Using the summation formulas $\sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6}$ and $\sum_{n=1}^{k} n = \frac{k(k+1)}{2}$:$S = 4 \sum_{n=1}^{13} n^2 - 4 \sum_{n=1}^{13} n + \sum_{n=1}^{13} 1$$S = 4 \left[ \frac{13(13+1)(2 	imes 13 + 1)}{6} ight] - 4 \left[ \frac{13(13+1)}{2} ight] + 13$$S = 4 \left[ \frac{13 	imes 14 	imes 27}{6} ight] - 2 	imes 13 	imes 14 + 13$$S = 4 	imes 13 	imes 7 	imes 9 - 364 + 13$$S = 3276 - 364 + 13 = 2925$.

The value of ${1^2} + {3^2} + {5^2} + \dots + {25^2}$ is

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