The sum of the series 1^2 + 2(2^2) + 3^2 + 2( | vedclass

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(A) The given series is $S = (1^2 + 3^2 + 5^2 + \dots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \dots + (2m)^2)$.Sum of odd squares up to $(2m-1)^2$ is $\sum_

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$m(2m+1)^2$

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(A) The given series is $S = (1^2 + 3^2 + 5^2 + \dots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \dots + (2m)^2)$.Sum of odd squares up to $(2m-1)^2$ is $\sum_{k=1}^{m} (2k-1)^2 = \frac{m(2m-1)(2m+1)}{3}$.Sum of even squares up to $(2m)^2$ is $2 	imes \sum_{k=1}^{m} (2k)^2 = 8 \sum_{k=1}^{m} k^2 = 8 	imes \frac{m(m+1)(2m+1)}{6} = \frac{4m(m+1)(2m+1)}{3}$.Adding these,$S = \frac{m(2m+1)}{3} [ (2m-1) + 4(m+1) ] = \frac{m(2m+1)}{3} [ 2m - 1 + 4m + 4 ] = \frac{m(2m+1)(6m+3)}{3} = m(2m+1)(2m+1) = m(2m+1)^2$.

The sum of the series $1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$ is

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