If S is the sum of n terms of the series 1 + | vedclass

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(A) The $n^{th}$ term of the series is given by $t_n = \frac{1 + 2 + 3 + \dots + n}{n}$.Using the formula for the sum of the first $n$ natural numbers

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$\frac{n(n + 3)}{4}$

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(A) The $n^{th}$ term of the series is given by $t_n = \frac{1 + 2 + 3 + \dots + n}{n}$.Using the formula for the sum of the first $n$ natural numbers,$\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}$,we get:$t_n = \frac{n(n + 1)}{2n} = \frac{1}{2}(n + 1)$.Now,the sum of $n$ terms $S_n$ is $\sum_{k=1}^{n} t_k = \sum_{k=1}^{n} \frac{1}{2}(k + 1)$.$S_n = \frac{1}{2} \left( \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \right)$.$S_n = \frac{1}{2} \left[ \frac{n(n + 1)}{2} + n \right]$.$S_n = \frac{n}{2} \left( \frac{n + 1}{2} + 1 \right) = \frac{n}{2} \left( \frac{n + 1 + 2}{2} \right) = \frac{n(n + 3)}{4}$.

If $S$ is the sum of $n$ terms of the series $1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \dots$,then $S = \dots$

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