If S_n = frac{n(n + 1)(n + 2)}{6},then sum_{n | vedclass

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(C) We know that the $n^{th}$ term $t_n = S_n - S_{n-1}$.Given $S_n = \frac{n(n + 1)(n + 2)}{6}$.$t_n = \frac{n(n + 1)(n + 2)}{6} - \frac{(n - 1)n(n +

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$2$

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(C) We know that the $n^{th}$ term $t_n = S_n - S_{n-1}$.Given $S_n = \frac{n(n + 1)(n + 2)}{6}$.$t_n = \frac{n(n + 1)(n + 2)}{6} - \frac{(n - 1)n(n + 1)}{6}$.Factoring out $\frac{n(n + 1)}{6}$,we get $t_n = \frac{n(n + 1)}{6} [n + 2 - (n - 1)]$.$t_n = \frac{n(n + 1)}{6} [3] = \frac{n(n + 1)}{2}$.Now,we need to find $\sum_{n = 1}^\infty \frac{1}{t_n} = \sum_{n = 1}^\infty \frac{2}{n(n + 1)}$.Using partial fractions,$\frac{2}{n(n + 1)} = 2 \left( \frac{1}{n} - \frac{1}{n + 1} \right)$.This is a telescoping series: $2 \left[ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots \right]$.The sum equals $2(1) = 2$.

If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

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