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Question

C01+C23+C45+C67......=

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Solution

We know that
(1+x)n=Cn0+Cn1x+Cn2x2+Cn3x3.......+Cnnxn(1)(1x)n=Cn0Cn1x+Cn2x2Cn3x3.......+(1)nCnnxn(2)
Integrating both sides of eq(1) under limits 0 to 1 we get
10(1+x)ndx=Cn010dx+Cn110xdx+......+Cnn10xndx(1+x)n+1n+110=Cn0+Cn12+Cn23+......+Cnnn+1(3)
Integrating both sides of eq(2) under limits 0 to 1 we get
10(1x)ndx=Cn010dxCn110xdx+......(1)n10Cnnxndx(1x)n+1n+110=Cn0Cn12+Cn23+......(1)nCnnn+1(4)
Adding (3) & (4) we get
(1+x)n+1n+110(1x)n+1n+110=2(Cn0+Cn23+Cn45.......)2n+11n+1+1n+1=2(Cn0+Cn23+Cn45.....)2n+1n+1=2(Cn0+Cn23+Cn45.....)Cn01+Cn23+Cn45.......=2nn+1


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