Lf n-N- then -k-1-k1-1-nk-1-

The correct option is C n^2 Let (1 1n) be tTherefore #160; ∑ kt^k 1 #160; =1+2t+3t^2...∞ =(1 t)^ 2 =(1 (1 1n))^ 2 =(1n)^ ...

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Sum of Product of Binomial Coefficients

lf n∈N, the...

Question

lf $n \in N$ , then $\infty \sum k = 1 k (1 - \frac{1}{n})^{k - 1} =$

n (n - 1)

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n (n + 1)

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n^{2}

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(n + 1)^{2}

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Solution

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Similar questions

n \sum r = 1 \frac{r^{n} C_{r}}{^{n} C_{r - 1}}

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

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

lim n \to \infty ((\frac{n}{n})^{n} + (\frac{n - 1}{n})^{n} + . . + ({\frac{1}{n}}^{n}))

(1^{2} - a_{1}) + (2^{2} - a_{2}) + . . . + (n^{2} - a_{n}) = \frac{1}{3} n (n^{2} - 1)

a_{n} = n + 1

1^{2} + 2^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

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