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Question

If n>3 and a,bϵR, then the value of abn(a1)(b1)+n(n1)1.2(a2)(b2)......+(1)n(an)(bn) is equal to

A
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C
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D
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Solution

The correct option is D
Tk+1 = (1)k .nCk(ak)(bk)
=(1)k. nCk[abk(a+b)+k2]
Thus, the sum of series in (i)
nk0(1)k.nCk[abk(a+b)+k2]
=abnk0(1)k.nCk(a+b)nk0(1)kk.nCk+nk0(1)kk2.nCk
We know that
(1+x)n=nC0+nC1x+nC2x2+...+nCnxn...(ii)
Differentiating both sides w.r.t. x, we get
n(1+x)n1=1.nC1+2.nC2x+3.nC3x2+...+n.nCnxn1...(iii)
Multiplying it by x we get
n.x(1+x)n1=1.nC1x+2.nC2x2+3.nC3x3+...+n.nCnxn
Differentiating w.r.t. x, we get
n(1+x)n1+n(n1)x(1+x)n2
=12.nC1+22.nC2x+...+n2.nCnxn1...(iv)
Putting x = –1 in (ii), (iii) and (iv), we get
nk0(1)k.nCk=0
nk0(1)kk.nCk=0
nk0(1)kk2.nCk=0
Thus, the sum of the series is 0

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