The sum of 24 terms of the following series s | vedclass

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(B) The given series is $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots$This can be rewritten as $1\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} +

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$300\sqrt{2}$

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(B) The given series is $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots$This can be rewritten as $1\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} + \dots$This is an arithmetic progression where the $n$-th term is $a_n = n\sqrt{2}$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} k\sqrt{2} = \sqrt{2} \sum_{k=1}^{n} k$.Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.For $n = 24$,the sum is $S_{24} = \sqrt{2} 	imes \frac{24 	imes 25}{2}$.$S_{24} = \sqrt{2} 	imes 12 	imes 25 = 300\sqrt{2}$.

The sum of $24$ terms of the following series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots$ is

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