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

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(C) We know that $t_n = S_n - S_{n-1}$.$t_n = \frac{n(n+1)(n+2)}{6} - \frac{(n-1)n(n+1)}{6}$$t_n = \frac{n(n+1)}{6} [ (n+2) - (n-1) ] = \frac{n(n+1)}{

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$2$

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(C) We know that $t_n = S_n - S_{n-1}$.$t_n = \frac{n(n+1)(n+2)}{6} - \frac{(n-1)n(n+1)}{6}$$t_n = \frac{n(n+1)}{6} [ (n+2) - (n-1) ] = \frac{n(n+1)}{6} 	imes 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,$\frac{2}{n(n+1)} = 2 \left( \frac{1}{n} - \frac{1}{n+1} ight)$.The sum is $2 \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} ight) = 2 [ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots ]$.This is a telescoping series,so the sum is $2 	imes 1 = 2$.

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

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