simplify Σ_(k=0) ^n (C


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Question Number 36437 by prof Abdo imad last updated on 02/Jun/18

$s i m p l i f y \sum_{k = 0}^{n} \frac{C_{n}^{k}}{k + 1}$
Commented by prakash jain last updated on 02/Jun/18

$\frac{1}{k + 1}^{n} C_{k} = (\frac{1}{k + 1}) \frac{n!}{(n - k)! k!}$ $= (\frac{1}{(n + 1)}) \frac{(n + 1)!}{(n - k)! (k + 1)!}$ $\underset{k = 0}{\sum^{n}}^{n} C_{k} = \frac{1}{n + 1} \underset{k = 0}{\sum^{n}}^{n + 1} C_{k + 1}$ $= \frac{1}{n + 1} (2^{n + 1} - 1)$
Commented by prof Abdo imad last updated on 03/Jun/18
$let consider the polynom P(x) =Σ_(k=0) ^n C_n ^k (x^(k+1) /(k+1)) we have (dP/dx)(x)= Σ_(k=0) ^n C_n ^k x^k = (x+1)^n ⇒ P(x) = ∫ (x+1)^n +c = (1/(n+1))(x+1)^(n+1) +c P(0)=0 = (1/(n+1)) +c ⇒c =−(1/(n+1)) ⇒ P(x) = (1/(n+1)){ (x+1)^n −1} and Σ_(k=0) ^n (C_n ^k /(k+1)) = P(1) = ((2^n −1)/(n+1 )) .$
$l e t c o n s i d e r t h e p o l y n o m P (x) = \sum_{k = 0}^{n} C_{n}^{k} \frac{x^{k + 1}}{k + 1}$ $w e h a v e \frac{d P}{d x} (x) = \sum_{k = 0}^{n} C_{n}^{k} x^{k} = {(x + 1)}^{n} \Rightarrow$ $P (x) = \int {(x + 1)}^{n} + c = \frac{1}{n + 1} {(x + 1)}^{n + 1} + c$ $P (0) = 0 = \frac{1}{n + 1} + c \Rightarrow c = - \frac{1}{n + 1} \Rightarrow$ $P (x) = \frac{1}{n + 1} {{(x + 1)}^{n} - 1} a n d$ $\sum_{k = 0}^{n} \frac{C_{n}^{k}}{k + 1} = P (1) = \frac{2^{n} - 1}{n + 1} .$
Commented by prof Abdo imad last updated on 03/Jun/18
$error at the final lines P(x)=(1/(n+1)){ (x+1)^(n+1) −1} and Σ_(k=0) ^n (C_n ^k /(k+1)) =P(1) = ((2^(n+1) −1)/(n+1)) ★$
$e r r o r a t t h e f i n a l l i n e s$ $P (x) = \frac{1}{n + 1} {{(x + 1)}^{n + 1} - 1} a n d$ $\sum_{k = 0}^{n} \frac{C_{n}^{k}}{k + 1} = P (1) = \frac{2^{n + 1} - 1}{n + 1} ★$
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