What is the sum of the series 1^2 + 2.2^2 + 3 | vedclass

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(D) The given series is $S = 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + (2m-1)^2 + (2m)^2$. This can be split into two series: $S_1 = 1^2 + 3^2

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$m(2m + 1)^2$

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(D) The given series is $S = 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + (2m-1)^2 + (2m)^2$. This can be split into two series: $S_1 = 1^2 + 3^2 + 5^2 + \dots + (2m-1)^2$ $S_2 = 2^2 + 4^2 + 6^2 + \dots + (2m)^2 = 2^2(1^2 + 2^2 + 3^2 + \dots + m^2) = 4 	imes \frac{m(m+1)(2m+1)}{6} = \frac{2m(m+1)(2m+1)}{3}$. For $S_1$,the sum of squares of first $m$ odd numbers is $\frac{m(2m-1)(2m+1)}{3}$. Adding both: $S = \frac{m(2m+1)}{3} [ (2m-1) + 2(m+1) ] = \frac{m(2m+1)}{3} [ 2m - 1 + 2m + 2 ] = \frac{m(2m+1)(4m+1)}{3}$. Wait,re-evaluating the series structure: $1^2 + 2^2 + 3^2 + \dots + (2m)^2 = \frac{2m(2m+1)(4m+1)}{6} = \frac{m(2m+1)(4m+1)}{3}$.

What is the sum of the series $1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2 + \dots + 2(2m)^2$?

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