(C1/C0) + (2C2 /C1) + ( 3C​3/ C2) +.... + nCn/Cn-1​= ? Pls solve using summation method. Thanks

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

S=nC1nC0+2nC2nC1+3nC3nC2+......+nnCnnCn-1S=r=1nrnCrnCr-1 ;iNownCrnCr-1=n!r! .n-r!n!r-1! n-r+1!=r-1! .n-r+1!r! .n-r!=r-1! .n-r+1.n-r!r.r-1!.n-r!=n-r+1rHence i becomes:S=r=1nrn-r+1r=r=1nn-r+1=r=1nn+r=1n1-r=1nr=nr=1n1+r=1n1-r=1nr=n1+1+1... n times+1+1+1... n times-1+2+3+.....+n=nn+n-nn+12  Sum of first n natural numbers=nn+12=n2+n-n2+n2=12n2+nnC1nC0+2nC2nC1+3nC3nC2+......+nnCnnCn-1=12n2+nPoint to remember:nCrnCr-1= n-r+1r

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