The sum to 20 terms of the series 2^2-3^2+4^2 | vedclass

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(C) The given series is $2^2-3^2+4^2-5^2+6^2-\ldots$ up to $20$ terms.This can be written as the sum of two series: $S = (2^2+4^2+6^2+\ldots 	ext{ to

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$1310$

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(C) The given series is $2^2-3^2+4^2-5^2+6^2-\ldots$ up to $20$ terms.This can be written as the sum of two series: $S = (2^2+4^2+6^2+\ldots 	ext{ to } 10 	ext{ terms}) - (3^2+5^2+7^2+\ldots 	ext{ to } 10 	ext{ terms})$.$S = \sum_{n=1}^{10} (2n)^2 - \sum_{n=1}^{10} (2n+1)^2$.$S = \sum_{n=1}^{10} [ (2n)^2 - (2n+1)^2 ]$.Using the identity $a^2-b^2 = (a-b)(a+b)$:$S = \sum_{n=1}^{10} (2n - 2n - 1)(2n + 2n + 1) = \sum_{n=1}^{10} (-1)(4n+1)$.$S = -[ 4 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 1 ]$.$S = -[ 4 	imes \frac{10 	imes 11}{2} + 10 ]$.$S = -[ 220 + 10 ] = -230$.

The sum to $20$ terms of the series $2^2-3^2+4^2-5^2+6^2-\ldots$ is equal to $........$.

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