1-k-1-C-n-k-

Question Number 141614 by ArielVyny last updated on 21/May/21

Σ \frac{1}{k + 1} C_{n}^{k} .

Commented by Dwaipayan Shikari last updated on 21/May/21

\frac{1}{k + 1} \underset{n = 0}{\sum^{k}} C_{n}^{k} = \frac{1}{k + 1} \underset{n = 0}{\sum^{k}} C_{n}^{k} {(1)}^{k - n} {(1)}^{n} = \frac{{(1 + 1)}^{k}}{k + 1} = \frac{2^{k}}{k + 1}

Commented by mathmax by abdo last updated on 21/May/21

not correct sir \dots

Answered by mathmax by abdo last updated on 21/May/21

let f (x) = \sum_{k = 0}^{n} \frac{1}{k + 1} C_{k}^{n} x^{k + 1} we have

f^{^{'}} (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} = {(x + 1)}^{n} \Rightarrow f (x) = \int {(x + 1)}^{n} dx + C

= \frac{{(x + 1)}^{n + 1}}{n + 1} + C

f (1) = \sum_{k = 0}^{n} \frac{C_{n}^{k}}{k + 1} = \frac{2^{n + 1}}{n + 1} + C

we f (0) = \frac{1}{n + 1} + C = 0 \Rightarrow C = - \frac{1}{n + 1} \Rightarrow \sum_{k = 0}^{n} \frac{C_{n}^{k}}{k + 1} = \frac{2^{n + 1}}{n + 1} - \frac{1}{n + 1}

= \frac{2^{n + 1} - 1}{n + 1}

Commented by ArielVyny last updated on 22/May/21

t h a n k s i r