The sum of (n + 1) terms of frac{1}{1} + frac | vedclass

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(D) The $k$-th term of the series is given by $T_k = \frac{1}{1 + 2 + 3 + \dots + k} = \frac{1}{\frac{k(k + 1)}{2}} = \frac{2}{k(k + 1)}$.Using partia

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$\frac{2(n + 1)}{n + 2}$

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(D) The $k$-th term of the series is given by $T_k = \frac{1}{1 + 2 + 3 + \dots + k} = \frac{1}{\frac{k(k + 1)}{2}} = \frac{2}{k(k + 1)}$.Using partial fractions,we can write $T_k = 2 \left[ \frac{1}{k} - \frac{1}{k + 1} \right]$.The sum of $(n + 1)$ terms is $S_{n + 1} = \sum_{k = 1}^{n + 1} T_k = \sum_{k = 1}^{n + 1} 2 \left[ \frac{1}{k} - \frac{1}{k + 1} \right]$.Expanding the summation: $S_{n + 1} = 2 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n + 1} - \frac{1}{n + 2}) \right]$.This is a telescoping series where intermediate terms cancel out,leaving $S_{n + 1} = 2 \left[ 1 - \frac{1}{n + 2} \right]$.Simplifying the expression: $S_{n + 1} = 2 \left[ \frac{n + 2 - 1}{n + 2} \right] = \frac{2(n + 1)}{n + 2}$.

The sum of $(n + 1)$ terms of $\frac{1}{1} + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots$ is:

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