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Question

C02+C12+C22+C32+.......?

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Solution

C20+C21+C33+........?(1+n)n=nC0+nC1x+nC2x2+......+nCnxn(x+1)n=nC0xn+nC1xn1+nCxxn2+...+nCn(1+x)2n=[nC0+nC1x+nC2x2+......+nCnxn]×[nC0xn+nC1xn1+nCxxn2+...+nCn]Coefficientofxn:2nCn=nC20+nC21+nC33+........+nCn.

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