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Question

Let Ck=nCk for 0kn and Ak=[C2k100C2k] for k1 and A1+A2+....An=[k100k2] then

A
k1=k2
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B
k1+k2=2nC2n+1
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C
k1=2nCn1
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D
k2=2nCn1
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Solution

The correct option is A k1=k2
WehaveAk=c2k100c22An=[c2000c21]+[c2100c22]+.....+[c2n100c2n]=[c20+c21+...+c2n11pt00c21+c22+...+c2n1pt]=[2ncn1002ncn1](1+x)n=nc0+nc1x+nc2x2+......+ncnxn(x+1)n=nc0xn+nc1xn1+nc2xx2+.......+ncn(1+x)2n=[nc0+nc1x+....+ncnxn][nc0xn+nc1xn1+....+ncn]coefficientofxn:2ncn=nc20+nc20+....+nc2nK1=K2=2ncn1Hence,optionAisthecorrectanswer.

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