The sum of all terms of the n^{th} bracket of | vedclass

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(D) The number of terms in successive groups are $1, 2, 3, \dots, n$. Thus,the $n^{th}$ group contains $n$ terms in an arithmetic progression with a c

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$n^3$

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(D) The number of terms in successive groups are $1, 2, 3, \dots, n$. Thus,the $n^{th}$ group contains $n$ terms in an arithmetic progression with a common difference $d = 2$.First,we find the first term of the $n^{th}$ group. Let $a_n$ be the first term of the $n^{th}$ group. The sequence of first terms is $1, 3, 7, 13, \dots$. The differences between consecutive terms are $2, 4, 6, \dots$,which form an arithmetic progression.Let $a_n = a_1 + \sum_{k=1}^{n-1} (2k) = 1 + 2 \cdot \frac{(n-1)n}{2} = 1 + n^2 - n = n^2 - n + 1$.The $n^{th}$ group is an arithmetic progression with $n$ terms,first term $a = n^2 - n + 1$,and common difference $d = 2$.The sum of the $n^{th}$ group is $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $S_n = \frac{n}{2} [2(n^2 - n + 1) + (n - 1)2] = \frac{n}{2} [2n^2 - 2n + 2 + 2n - 2] = \frac{n}{2} [2n^2] = n^3$.

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

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