The sum of the terms in the $n^{t h}$ bracket of the series $(1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) + \dots$ is

(n - 1)^{3} + n^{3}

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(n + 1)^{3} + 8 n^{2}

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\frac{(n + 1) (n + 2)}{6 n}

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(n + 1)^{3} + n^{3}

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Solution

The correct option is A $(n - 1)^{3} + n^{3}$
For $n = 1$ , we have
Sum of the terms in first bracket $=$ 1 and, $(n - 1)^{3} + n^{3} = (1 - 1)^{3} + 1^{3} = 1$
For $n = 2$ , we have
Sum of the terms in the second bracket $= 2 + 3 + 4 =$ 9 and, $(n - 1)^{3} + n^{3} = (2 - 1)^{3} + 2^{3} = 1 + 8 = 9$
For $n = 3$ , we have
Sum of the terms in the third bracket $= (5 + 6 + 7 + 8 + 9) = 35$ and, $(3 - 1)^{3} + 3^{3} = 8 + 27 = 35$
Hence, option (A) is true.

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