Menu Close

Prove-that-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-

Tinku Tara 0 Comments
Pin




Question Number 24142 by Tinkutara last updated on 13/Nov/17
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
Provethat∑2n−1r=1(−1)r−1(∫10xr(1−x)2n−rdx)=∫10[(1−x)2n+x2n−(1−x)2n+1−x2n+1]dx
Commented by Tinkutara last updated on 13/Nov/17
Thank you very much Sir!
ThankyouverymuchSir!
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)
⇔Show∑2n−1r=1(−1)r−1xr(1−x)2n−r=(1−x)2n+x2n−(1−x)2n+1−x2n+1LetS=∑2n−1r=1(−1)r−1xr(1−x)2n−r(1−x)S+xS=∑2n−1r=1(−1)r−1xr(1−x)2n−r+1+∑2n−1r=1(−1)r−1xr+1(1−x)2n−rS={x(1−x)2n+∑2n−1r=2(−1)r−1xr(1−x)2n−r+1}+{∑2n−2r=1(−1)r−1xr+1(1−x)2n−r+x2n(1−x)}S=x(1−x)2n+x2n(1−x)+∑2n−1r=2(−1)r−1xr(1−x)2n−r+1+{∑2n−1r=2(−1)r−2xr(1−x)2n−r+1}r→r−1S=x(1−x)2n+x2n(1−x)+∑2n−1r=2{(−1)r−1xr(1−x)2n−r+1+(−1)r−2xr(1−x)2n−r+1}S=x(1−x)2n+x2n(1−x)+∑2n−1r=2xr(1−x)2n−r+1{(−1)r−1+(−1)r−2}↑=0S=x(1−x)2n+x2n(1−x)S={−(1−x)+1}(1−x)2n+x2n(1−x)S=(1−x)2n−(1−x)2n+1+x2n−x2n+1

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *