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Question

If Ar,Br,Cr denotes the coefficients of xr in the expansion of (1+x)10,(x+1)20,(1+x)30 respectively, then the value of 10r=1Ar(B10BrC10Ar) is

A
B10C10
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B
A10(B210C10A10)
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C
0
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D
C10B10
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Solution

The correct option is D C10B10
Given series
10r=1Ar(B10BrC10Ar)=B1010r=1ArBrC1010r=1(Ar)2

Now,
(1+x)10=A0+A1x+A2x2++A10x10
(x+1)20=B0x20+B1x19+B2x18++B20

Now, coefficient of x20 in (1+x)10(x+1)20 is A0B0+A1B1+A2B2++A10B10

As (1+x)10(x+1)20=(1+x)30
So, coefficient of x20 is C20

10r=0ArBr=C20=C1010r=1ArBr=C101(1)

We know that
( nC0)2+( nC1)2++( nCn)2= 2nCnA20+A21+A22++A210= 20C1010r=1A2r=B101(2)

Therefore,
10r=1Ar(B10BrC10Ar)=B1010r=1ArBrC1010r=1A2r
Using equation (1) and (2), we get
=B10(C101)C10(B101)=C10B10

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