( C0/2 ) - ( C1/3) + (C2 /4) - ( C3/5) +...... =? Pls solve using summation method. Thanks.

Dear Student,
Please find below the solution to the asked query:

We have:S=nC02-nC13+nC24-nC25+.....S=r=0n -1rnCrr+2=r=0n -1r r+1r+2nCrr+1We know that:nCrr+1=n+1Cr+1n+1 S=r=0n -1r r+1r+2n+1Cr+1n+1=1n+1r=0n -1r r+1n+1Cr+1r+2=1n+1r=0n -1r r+1n+1Cr+1r+1+1Again applying same formula we get,S=1n+1r=0n -1r r+1n+2Cr+2n+2=1n+1n+2r=0n -1r r+1.n+2Cr+2=1n+1n+2r=0n -1r r+2-1.n+2Cr+2=1n+1n+2r=0n-1r r+2.n+2Cr+2-r=0n-1r .n+2Cr+2We know thatr.nCr=n.n-1Cr-1 r+2.n+2Cr+2=n+2.n+1Cr+1S=1n+1n+2n+2r=0n-1r .n+1Cr+1-r=0n-1r .n+2Cr+2 ir=0n-1r .n+1Cr+1=n+1C1-n+1C2+n+1C3-..........=-n+1C0+n+1C1-n+1C2+n+1C3-..........+n+1C0=n+1C1+n+1C3+...-n+1C0+n+1C2+...+n+1C0=0+n+1C0 Sum of odd Binomial Coefficient=Sum of Even Binomial Coefficient=n+1C0=1r=0n-1r .n+2Cr+2=n+2C2-n+2C3+n+2C4-n+2C5+...=n+2C0-n+2C1+n+2C2-n+2C3+n+2C4-n+2C5+...+n+2C1-n+2C0=0+n+2C1-n+2C0=n+2-1=n+1Hence i becomes:S=1n+1n+2n+2-n+1=1n+1n+2n+2-n-1=1n+1n+2nC02-nC13+nC24-nC25+.....=1n+1n+2

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