The sum of the series (2)^2 + 2(4)^2 + 3(6)^2 | vedclass

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(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms.The $n$-th term of the series is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.To fin

Vedclass · Accepted Answer

$12100$

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(C) The given series is $S = 1(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms.The $n$-th term of the series is $T_n = n(2n)^2 = n(4n^2) = 4n^3$.To find the sum of $10$ terms,we calculate $\sum_{n=1}^{10} 4n^3$.$S = 4 \sum_{n=1}^{10} n^3$.Using the formula for the sum of cubes of first $n$ natural numbers,$\sum_{n=1}^{n} n^3 = [\frac{n(n+1)}{2}]^2$.For $n = 10$,$\sum_{n=1}^{10} n^3 = [\frac{10 	imes 11}{2}]^2 = (55)^2 = 3025$.Therefore,$S = 4 	imes 3025 = 12100$.

The sum of the series $(2)^2 + 2(4)^2 + 3(6)^2 + ...$ up to $10$ terms is:

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