Question-7013

Question Number 7013 by Master Moon last updated on 06/Aug/16

Commented by Yozzis last updated on 06/Aug/16

{(1 - x)}^{n} = \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) {(- 1)}^{k} x^{k} = 1 + \underset{k = 1}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) {(- 1)}^{k} x^{k} (x \in R) {B i n o m i a l t h e o r e m)

\Rightarrow \int_{1}^{0} {(1 - x)}^{n} d x = \int_{1}^{0} {1 - (\begin{matrix} n \\ 1 \end{matrix}) x + (\begin{matrix} n \\ 2 \end{matrix}) x^{2} - (\begin{matrix} n \\ 3 \end{matrix}) x^{3} + \dots + (\begin{matrix} n \\ n \end{matrix}) {(- 1)}^{n} x^{n}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x

\frac{{(1 - x)}^{n + 1}}{n + 1} ∣_{0}^{1} = - {[x - \frac{1}{2} (\begin{matrix} n \\ 1 \end{matrix}) x^{2} + \frac{1}{3} (\begin{matrix} n \\ 2 \end{matrix}) x^{3} - \frac{1}{4} (\begin{matrix} n \\ 3 \end{matrix}) x^{4} + \dots + \frac{1}{n + 1} {(- 1)}^{n} (\begin{matrix} n \\ n \end{matrix}) x^{n + 1}]}_{0}^{1}

(\frac{1 - 1}{n + 1} - \frac{1^{n + 1}}{n + 1}) = - (1 - \frac{1}{2} (\begin{matrix} n \\ 1 \end{matrix}) + \frac{1}{3} (\begin{matrix} n \\ 3 \end{matrix}) - \frac{1}{4} (\begin{matrix} n \\ 4 \end{matrix}) + \dots + \frac{{(- 1)}^{n}}{n + 1} (\begin{matrix} n \\ n \end{matrix}))

- \frac{1}{n + 1} = - 1 + \frac{1}{2} (\begin{matrix} n \\ 1 \end{matrix}) - \frac{1}{3} (\begin{matrix} n \\ 2 \end{matrix}) + \frac{1}{4} (\begin{matrix} n \\ 3 \end{matrix}) + \dots + \frac{{(- 1)}^{n + 1}}{n + 1} (\begin{matrix} n \\ n \end{matrix})

\frac{1}{2} (\begin{matrix} n \\ 1 \end{matrix}) - \frac{1}{3} (\begin{matrix} n \\ 2 \end{matrix}) + \frac{1}{4} (\begin{matrix} n \\ 3 \end{matrix}) + \dots + \frac{{(- 1)}^{n + 1}}{n + 1} (\begin{matrix} n \\ n \end{matrix}) = \frac{n}{n + 1}

Answered by Yozzii last updated on 08/Aug/16

C h e c k f o r a n a n s w e r i n t h e c o m m e n t s .

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