Prove that Σ_(r=1) ^(2n−1) (−1)^(r−1) (∫_0 ^1 x^r (1−x)^(2n−r) dx) =∫_0 ^1 [(1−x)^(2n) +x^(2n) −(1−x)^(2n+1) −x^(2n+1) ]dx


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Question Number 24142 by Tinkutara last updated on 13/Nov/17

$P r o v e t h a t$ $\underset{r = 1}{\sum^{2 n - 1}} {(- 1)}^{r - 1} (\underset{0}{\int^{1}} x^{r} {(1 - x)}^{2 n - r} d x)$ $= \underset{0}{\int^{1}} [{(1 - x)}^{2 n} + x^{2 n} - {(1 - x)}^{2 n + 1} - x^{2 n + 1}] d x$
Commented by Tinkutara last updated on 13/Nov/17

$Thank you very much Sir!$
Commented by moxhix last updated on 13/Nov/17
$⇔ShowΣ_(r=1) ^(2n−1) (−1)^(r−1) x^r (1−x)^(2n−r) =(1−x)^(2n) +x^(2n) −(1−x)^(2n+1) −x^(2n+1) Let S=Σ_(r=1) ^(2n−1) (−1)^(r−1) x^r (1−x)^(2n−r) (1−x)S+xS=Σ_(r=1) ^(2n−1) (−1)^(r−1) x^r (1−x)^(2n−r+1) +Σ_(r=1) ^(2n−1) (−1)^(r−1) x^(r+1) (1−x)^(2n−r) S={x(1−x)^(2n) +Σ_(r=2) ^(2n−1) (−1)^(r−1) x^r (1−x)^(2n−r+1) }+{Σ_(r=1) ^(2n−2) (−1)^(r−1) x^(r+1) (1−x)^(2n−r) +x^(2n) (1−x)} S=x(1−x)^(2n) +x^(2n) (1−x)+Σ_(r=2) ^(2n−1) (−1)^(r−1) x^r (1−x)^(2n−r+1) +{Σ_(r=2) ^(2n−1) (−1)^(r−2) x^r (1−x)^(2n−r+1) }_(r→r−1) S=x(1−x)^(2n) +x^(2n) (1−x)+Σ_(r=2) ^(2n−1) {(−1)^(r−1) x^r (1−x)^(2n−r+1) +(−1)^(r−2) x^r (1−x)^(2n−r+1) } S=x(1−x)^(2n) +x^(2n) (1−x)+Σ_(r=2) ^(2n−1) x^r (1−x)^(2n−r+1) {(−1)^(r−1) +(−1)^(r−2) }_(↑=0) S=x(1−x)^(2n) +x^(2n) (1−x) S={−(1−x)+1}(1−x)^(2n) +x^(2n) (1−x) S=(1−x)^(2n) −(1−x)^(2n+1) +x^(2n) −x^(2n+1)$
$\Leftrightarrow S h o w \underset{r = 1}{\sum^{2 n - 1}} {(- 1)}^{r - 1} x^{r} {(1 - x)}^{2 n - r} = {(1 - x)}^{2 n} + x^{2 n} - {(1 - x)}^{2 n + 1} - x^{2 n + 1}$ $L e t S = \underset{r = 1}{\sum^{2 n - 1}} {(- 1)}^{r - 1} x^{r} {(1 - x)}^{2 n - r}$ $(1 - x) S + x S = \underset{r = 1}{\sum^{2 n - 1}} {(- 1)}^{r - 1} x^{r} {(1 - x)}^{2 n - r + 1} + \underset{r = 1}{\sum^{2 n - 1}} {(- 1)}^{r - 1} x^{r + 1} {(1 - x)}^{2 n - r}$ $S = {x {(1 - x)}^{2 n} + \underset{r = 2}{\sum^{2 n - 1}} {(- 1)}^{r - 1} x^{r} {(1 - x)}^{2 n - r + 1}} + {\underset{r = 1}{\sum^{2 n - 2}} {(- 1)}^{r - 1} x^{r + 1} {(1 - x)}^{2 n - r} + x^{2 n} (1 - x)}$ $S = x {(1 - x)}^{2 n} + x^{2 n} (1 - x) + \underset{r = 2}{\sum^{2 n - 1}} {(- 1)}^{r - 1} x^{r} {(1 - x)}^{2 n - r + 1} + \underset{r \to r - 1}{{\underset{r = 2}{\sum^{2 n - 1}} {(- 1)}^{r - 2} x^{r} {(1 - x)}^{2 n - r + 1}}}$ $S = x {(1 - x)}^{2 n} + x^{2 n} (1 - x) + \underset{r = 2}{\sum^{2 n - 1}} {{(- 1)}^{r - 1} x^{r} {(1 - x)}^{2 n - r + 1} + {(- 1)}^{r - 2} x^{r} {(1 - x)}^{2 n - r + 1}}$ $S = x {(1 - x)}^{2 n} + x^{2 n} (1 - x) + \underset{r = 2}{\sum^{2 n - 1}} x^{r} {(1 - x)}^{2 n - r + 1} \underset{↑= 0}{{{(- 1)}^{r - 1} + {(- 1)}^{r - 2}}}$ $S = x {(1 - x)}^{2 n} + x^{2 n} (1 - x)$ $S = {- (1 - x) + 1} {(1 - x)}^{2 n} + x^{2 n} (1 - x)$ $S = {(1 - x)}^{2 n} - {(1 - x)}^{2 n + 1} + x^{2 n} - x^{2 n + 1}$
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