If -r-1ntr-112nn-1n-2- then value of -r-1n1tr is

The correct option is D 4n(n+1) We have t_r=∑_k=1^rt_k ∑_k=1^r 1t_k 1=112r(r+1)(r+2) 112(r 1)(r)(r+1) =14r(r+1) Now, 1t_r=4r(r+1)=4(1r 1r+1 ...

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If ∑r=1ntr=...

Question

If $n \sum r = 1 t_{r} = \frac{1}{12} n (n + 1) (n + 2)$ , then value of $n \sum r = 1 \frac{1}{t_{r}}$ is

\frac{2 n}{n + 1}

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\frac{n - 1}{(n + 1)!}

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\frac{4 n}{(n + 1)}

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\frac{3 n}{n + 2}

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n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}

p

n \sum r = 1 t_{r} = \frac{1}{12} n (n + 1) (n + 2)

n \sum r = 1 \frac{1}{t_{r}}

n \sum r = 1 t_{r} = n \sum k = 1 k \sum j = 1 j \sum i = 1 (2)

n \sum r = 1 {\frac{1}{t}}_{r}

n \sum r = 1 T_{r} = \frac{n}{8} (n + 1) (n + 2) (n + 3)

n \sum r = 1 \frac{1}{T_{r}} = \frac{n^{2} + 3 n}{4 p \sum k = 1 k}

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

\sum_{r = 1}^{n} \frac{1}{T_{r}}

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