The n^{text{th}} term of the series 1 + (3 + | vedclass

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(D) The given series is $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ The number of terms in the $n^{	ext{th}}$ group is $2n - 1$. Let the fir

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$(2n - 1)[(n - 1)^2 + n^2]$

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(D) The given series is $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ The number of terms in the $n^{	ext{th}}$ group is $2n - 1$. Let the first term of the $n^{	ext{th}}$ group be $a_n$. The total number of terms before the $n^{	ext{th}}$ group is $1 + 3 + 5 + \ldots + (2(n-1) - 1) = (n-1)^2$. Thus,the first term of the $n^{	ext{th}}$ group is $a_n = (n-1)^2 + 1$. The $n^{	ext{th}}$ group is an arithmetic progression with $2n-1$ terms,first term $a_n$,and common difference $d = 2$. The sum of the $n^{	ext{th}}$ group $t_n$ is: $t_n = \frac{2n-1}{2} [2a_n + (2n-2)d] = (2n-1) [a_n + (n-1)2]$ $t_n = (2n-1) [(n-1)^2 + 1 + 2n - 2] = (2n-1) [n^2 - 2n + 1 + 1 + 2n - 2] = (2n-1) [n^2 + (n-1)^2]$.

The $n^{\text{th}}$ term of the series $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ is:

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