If ∑r=1n Tr=n8(n+1)(n+2)(n+3), find ∑r=1n 1Tr

Sn=∑r=1n Tr=n(n+1)(n+2)(n+3)8Tr=Sr−Sr−1=r(r+1)(r+2)(r+3)8−(r−1)r(r+1)(r+2)8=r(r+1)(r+2)21Tr=2r(r+1)(r+2)=(r+2)−rr(r+1)(r+2)=1r(r+1)−1(r+1)(r+2)∑r=1n 1Tr=∑r=1n 1r(r+1)−1(r+1)(r+2)=∑r=1n 1r−1r+1−1r+1−1r+2=11−1n+1−12−1n+2=12+1n+2−1n+1=n(n+3)2(n+1)(n+2)

If ∑r=1n Tr=n8(n+1)(n+2)(n+3), find ∑r=1n 1Tr

Sn=∑r=1n Tr=n(n+1)(n+2)(n+3)8Tr=Sr−Sr−1=r(r+1)(r+2)(r+3)8−(r−1)r(r+1)(r+2)8=r(r+1)(r+2)21Tr=2r(r+1)(r+2)=(r+2)−rr(r+1)(r+2)=1r(r+1)−1(r+1)(r+2)∑r=1n 1Tr=∑r=1n 1r(r+1)−1(r+1)(r+2)=∑r=1n 1r−1r+1−1r+1−1r+2=11−1n+1−12−1n+2=12+1n+2−1n+1=n(n+3)2(n+1)(n+2)

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If $\sum_{r = 1}^{n} T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)$ , find $\sum_{r = 1}^{n} \frac{1}{T_{r}}$

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$\frac{(n + 3)}{(n + 1) (n + 2)}$

$\frac{n (n + 3)}{2 (n + 1) (n + 2)}$

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

$\frac{n (n + 3)}{(n + 1) (n + 2)}$

answer is B.

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Detailed Solution

$\begin{array}{l} S_{n} = \sum_{r = 1}^{n} T_{r} = \frac{n (n + 1) (n + 2) (n + 3)}{8} \\ T_{r} = S_{r} - S_{r - 1} = \frac{r (r + 1) (r + 2) (r + 3)}{8} - \frac{(r - 1) r (r + 1) (r + 2)}{8} \\ = \frac{r (r + 1) (r + 2)}{2} \\ \frac{1}{T_{r}} = \frac{2}{r (r + 1) (r + 2)} = \frac{(r + 2) - r}{r (r + 1) (r + 2)} = (\frac{1}{r (r + 1)} - \frac{1}{(r + 1) (r + 2)}) \end{array}$

$\begin{array}{l} \sum_{r = 1}^{n} \frac{1}{T_{r}} = \sum_{r = 1}^{n} (\frac{1}{r (r + 1)} - \frac{1}{(r + 1) (r + 2)}) \\ = \sum_{r = 1}^{n} \{(\frac{1}{r} - \frac{1}{r + 1}) - (\frac{1}{r + 1} - \frac{1}{r + 2})\} \\ = (\frac{1}{1} - \frac{1}{n + 1}) - (\frac{1}{2} - \frac{1}{n + 2}) \\ = \frac{1}{2} + \frac{1}{n + 2} - \frac{1}{n + 1} = \frac{n (n + 3)}{2 (n + 1) (n + 2)} \end{array}$