Let-1-x-n-r-0-n-C-r-x-r-and-r-0-n-C-r-r-1-k-then-the-value-of-k-is-

Question Number 63942 by gunawan last updated on 11/Jul/19

Let {(1 + x)}^{n} = \underset{r = 0}{\sum^{n}} C_{r} x^{r} and \underset{r = 0}{\sum^{n}} \frac{C_{r}}{r + 1} = k,

then the value of k is

Commented by mathmax by abdo last updated on 11/Jul/19

{(1 + x)}^{n} = \sum_{r = 0}^{n} C_{n}^{r} x^{r} \Rightarrow C_{r} = C_{n}^{r} \Rightarrow

\sum_{r = 0}^{n} \frac{C_{r}}{r + 1} = \sum_{r = 0}^{n} \frac{C_{n}^{r}}{r + 1}

l e t p (x) = \sum_{r = 0}^{n} \frac{C_{n}^{r}}{r + 1} x^{r + 1} \Rightarrow p^{^{'}} (x) = \sum_{r = 0}^{n} C_{n}^{r} x^{r} = {(x + 1)}^{n} \Rightarrow

p (x) = \int {(x + 1)}^{n} d x + c = \frac{1}{n + 1} {(x + 1)}^{n + 1} + c

p (0) = 0 = \frac{1}{n + 1} + c \Rightarrow c = - \frac{1}{n + 1} \Rightarrow p (x) = \frac{{(x + 1)}^{n + 1} - 1}{n + 1} \Rightarrow

\sum_{r = 0}^{\infty} \frac{C_{r}}{r + 1} = p (1) = \frac{2^{n + 1} - 1}{n + 1} \Rightarrow k = \frac{2^{n + 1} - 1}{n + 1} .

Answered by mr W last updated on 11/Jul/19

{(1 + x)}^{n} = \underset{r = 0}{\sum^{n}} C_{r} x^{r}

\int_{0}^{x} {(1 + x)}^{n} d x = \underset{r = 0}{\sum^{n}} \int_{0}^{x} C_{r} x^{r} d x

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

l e t x = 1 :

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

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

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