Find the sum of the series for the first $n$ brackets $(1)$ + $(2 + 3 + 4) + (5 + 6 + 7 + 8 + 9) +$ ... 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}

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\frac{(n + 3) (n + 2)}{12}

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Solution

The correct option is C $(n - 1)^{3} + n^{3}$
All the brackets are in A.P with initial term being $(n - 1)^{2} + 1$ and the final term being $n^{2}$
Number of terms in a bracket $= n^{2} - (n - 1)^{2}$
$= 2 n - 1$
Therefore Sum of an AP $= \frac{n}{2} (a + l)$
where,
$n :$ no of terms
$a :$ first term
$l :$ last term
$= \frac{2 n - 1}{2} \times ((n - 1)^{2} + 1 + n^{2})$
$= (2 n - 1) (n^{2} + 1 - n)$
$= (n - 1)^{3} + n^{3}$

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