For r=0,1,...,10, let Ar,Br, and Cr denote, respctively, the coefficient of xr in the expansions of (1+x)10,(1+x)20 and (1+x)30. Then ∑10r=1Ar(B10Br−C10Ar)is equal to
A
B10−C10
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B
A10(B210C10A10)
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C
\N
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D
C10−B10
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Solution
The correct option is DC10−B10 ClearlyAr=10Cr,Br=20Cr,Cr=30Cr,Now∑10r=110Cr(20C1020Cr−30C1010Cr)=20C10∑10r=110Cr20Cr−30C10∑10r=110Cr×10Cr=20C10(10C120C1+10C220C2+...+10C1020C10)=30C10(10C1×10C1+10C2×10C2+...+10C1010C10)...(1)
Now expanding (1+x)10 and (1+x)20 by binomial theorem and comparing the coefficients of x20 in their product, on both sides, we get 10C020C0+10C120C1+10C220C2+...+10C1020C10
= coeff of x20 in (1+x)30=30C20=30C10∴10C120C1+10C220C2+...+10C1020C10=30C10−1
Again expending (1+x10) and (x+1)10 by binomial theorem and comparing the coefficients of x10 in their product on both sides, we get ∴(10C0)2(10C1)2+(10C2)2+...+(10C10)2
= coeff of x10 in (1+x)20=20C10∴(10C1)2+(10C2)2+...+(10C10)2=20C10−1
Substituting these values in equation (1), we get 20C10(30C10−1)−30C10(20C10−1)=30C10−20C10=C10−B10