The sum of the series
$(2)^{2} + 2 (4)^{2} + 3 (6)^{2} + . . . .$ upto $10$ terms is

12100

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11300

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11200

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12300

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Solution

The correct option is A $12100$
${(2)}^{2} + 2 {(4)}^{2} + 3 {(6)}^{2} + \dots . . u p t o 10 t e r m$
$T_{m} = n {(2 n)}^{2} \Rightarrow 4 n^{3}$
$\sum_{n = 0}^{10} T_{n} = 4 \sum_{n = 0}^{10} n^{3} = \frac{4 {(n)}^{2} {(n + 1)}^{2}}{4}$
$n = 10 \to= 100 \times 121 = 12100$

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