If an = Σ1/nCr , then Σr/nCr equals where summation is from r=0 to r=n.

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Please find below the solution to the asked query:

an=r=0n 1 nCr=1nC0+1nC1+1nC2+1nC3+.....+1nCn-3+1nCn-2+1nCn-1+1nCn=1+1nC1+1nC2+1nC3+.....+1nC3+1nC2+1nC1+1 As nCr=nCn-r and nCn=nC0=1an=2+21nC1+1nC2+1nC3+an-22=1nC1+1nC2+1nC3+...iS=r=0n r nCr=0+1nC1+2nC2+3nC3+.....+n-3nCn-3+n-2nCn-2+n-1nCn-1+nnCn=1nC1+2nC2+3nC3+.....+n-3nC3+n-2nC2+n-1nC1+nnCn  As nCr=nCn-r=1+n-1nC1+2+n-2nC2+3+n-3nC3+... nnCn=nnC1+nnC2+nnC3+... nnCn=n1nC1+1nC2+1nC3+..+... n1=nan-22+n Using i=nan-22+1=nan-2+22=n an2r=0n r nCr=n an2

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