Value of L-limn- n14-1-k-1nk-2-k-1n-1k-3-k-1n-2k-n-1- is

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Value of L=...

Question

Value of $L = lim n \to \infty n \frac{1}{4} [1. (n \sum k = 1 k) + 2. (n - 1 \sum k = 1 k) + 3. (n - 2 \sum k = 1 k) + . . . + n .1]$ is

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L = lim n \to \infty \frac{1}{n^{4}} [1 (n \sum k = 1 k) + 2 (n - 1 \sum k = 1 k) + 3 (n - 2 \sum k = 1 k) + . . . . . . + n .1]

L = lim n \to \infty \frac{1}{n^{4}} [1. (n \sum k = 1 k) + 2. (n \sum k = 1 k) + 3. (n \sum k = 1 k) + . . . . . . n .1]

$\infty \sum n = 1 \frac{1}{(n + 1) (n + 2) (n + 3) . . . . (n + k)}$ is equal to

2^{k} (\frac{n}{0}) (\frac{n}{k}) - 2^{k - 1} (\frac{n}{1}) (\frac{n - 1}{k - 1}) + 2^{k - 2} (\frac{n}{2}) (\frac{n - 2}{k - 2}) . . + (- 1)^{k} (\frac{n}{k}) (\frac{n - k}{0})

lim n \to \infty n [\frac{1}{(n + 1) (n + 2)} + \frac{1}{(n + 2) (n + 4)} + \dots + \frac{1}{6 n^{2}}] = log k

2 k =

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