The sum of the 1^{st} n terms of the series f | vedclass

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(B) The $n^{th}$ term of the series is given by $t_n = \frac{1^2 + 2^2 + \ldots + n^2}{1 + 2 + \ldots + n}$.Using the formulas for the sum of the firs

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

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(B) The $n^{th}$ term of the series is given by $t_n = \frac{1^2 + 2^2 + \ldots + n^2}{1 + 2 + \ldots + n}$.Using the formulas for the sum of the first $n$ natural numbers and their squares:$t_n = \frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{2n+1}{3}$.Now,the sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} \frac{2k+1}{3}$.$S_n = \frac{1}{3} [2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1]$.$S_n = \frac{1}{3} [2 \cdot \frac{n(n+1)}{2} + n] = \frac{1}{3} [n(n+1) + n] = \frac{1}{3} [n^2 + n + n] = \frac{n(n+2)}{3}$.

The sum of the $1^{st} n$ terms of the series $\frac{1^{2}}{1} + \frac{1^{2}+2^{2}}{1+2} + \frac{1^{2}+2^{2}+3^{2}}{1+2+3} + \ldots$ is:

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