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

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(A) The given series is $S = 1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$.We can group the odd and even terms:$S = (1^2 + 3^2 + 5^2 +

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$m(2m+1)^2$

Vedclass · Answer

(A) The given series is $S = 1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$.We can group the odd and even terms:$S = (1^2 + 3^2 + 5^2 + \dots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \dots + (2m)^2)$.$S = \sum_{k=1}^{m} (2k-1)^2 + 2 \sum_{k=1}^{m} (2k)^2$.$S = \sum_{k=1}^{m} (4k^2 - 4k + 1) + 8 \sum_{k=1}^{m} k^2$.$S = 4 \sum k^2 - 4 \sum k + \sum 1 + 8 \sum k^2 = 12 \sum_{k=1}^{m} k^2 - 4 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 1$.Using standard summation formulas:$S = 12 \frac{m(m+1)(2m+1)}{6} - 4 \frac{m(m+1)}{2} + m$.$S = 2m(m+1)(2m+1) - 2m(m+1) + m$.$S = m [2(m+1)(2m+1) - 2(m+1) + 1]$.$S = m [2(2m^2 + 3m + 1) - 2m - 2 + 1]$.$S = m [4m^2 + 6m + 2 - 2m - 1] = m(4m^2 + 4m + 1) = m(2m+1)^2$.

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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