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 $A$.$P$. with a common differen

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