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(A) The $n^{th}$ group contains $n$ terms in an arithmetic progression with a common difference of $1$.The first term of the $n^{th}$ group,$T_n$,is t

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$\frac{1}{2}\,[n\,({n^2} + 1)]$

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(A) The $n^{th}$ group contains $n$ terms in an arithmetic progression with a common difference of $1$.The first term of the $n^{th}$ group,$T_n$,is the sum of the number of terms in the previous $(n-1)$ groups plus $1$.The number of terms in the first $(n-1)$ groups is $1 + 2 + 3 + \dots + (n-1) = \frac{(n-1)n}{2}$.Thus,the first term of the $n^{th}$ group is $T_n = \frac{n(n-1)}{2} + 1 = \frac{n^2 - n + 2}{2}$.The sum of the $n$ terms in the $n^{th}$ group is given by the formula for the sum of an arithmetic progression: $S_n = \frac{n}{2} [2a + (n-1)d]$,where $a = T_n$ and $d = 1$.$S_n = \frac{n}{2} [2(\frac{n^2 - n + 2}{2}) + (n-1)(1)]$$S_n = \frac{n}{2} [n^2 - n + 2 + n - 1]$$S_n = \frac{n}{2} [n^2 + 1] = \frac{1}{2} [n(n^2 + 1)]$.

The sequence of natural numbers is divided into groups as follows: $(1), (2, 3), (4, 5, 6), (7, 8, 9, 10), \dots$. Find the sum of the numbers in the $n^{th}$ group.

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