If n>3 and a,bϵR, then the value of ab−n(a−1)(b−1)+n(n−1)1.2(a−2)(b−2)−......+(−1)n(a−n)(b−n) is equal to
A
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B
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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(a−k)(b−k) =(−1)k.nCk[ab−k(a+b)+k2] Thus, the sum of series in (i) ∑nk−0(−1)k.nCk[ab−k(a+b)+k2] =ab∑nk−0(−1)k.nCk−(a+b)∑nk−0(−1)kk.nCk+∑nk−0(−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)n−1=1.nC1+2.nC2x+3.nC3x2+...+n.nCnxn−1...(iii) Multiplying it by x we get n.x(1+x)n−1=1.nC1x+2.nC2x2+3.nC3x3+...+n.nCnxn Differentiating w.r.t. x, we get n(1+x)n−1+n(n−1)x(1+x)n−2 =12.nC1+22.nC2x+...+n2.nCnxn−1...(iv) Putting x = –1 in (ii), (iii) and (iv), we get ∑nk−0(−1)k.nCk=0 ∑nk−0(−1)kk.nCk=0 ∑nk−0(−1)kk2.nCk=0 Thus, the sum of the series is 0