If the sum of 1 + frac{1 + 2}{2} + frac{1 + 2 | 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,$\sum n = \frac{n(n + 1)}{2}$,we get:$T_n = \frac{n(n + 1)}{2n} = \frac{n + 1}{2}$.Now,the sum $S$ of $n$ terms is $S = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k + 1}{2}$.$S = \frac{1}{2} \left( \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \right)$.$S = \frac{1}{2} \left( \frac{n(n + 1)}{2} + n \right)$.$S = \frac{1}{2} \left( \frac{n^2 + n + 2n}{2} \right) = \frac{n^2 + 3n}{4} = \frac{n(n + 3)}{4}$.

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

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