Binomial TheoremHard
Question
The sum of the series 
(-1)r-1 . nCr(a - r) is equal to :
(-1)r-1 . nCr(a - r) is equal to :Options
A.n . 2n - 1 + a
B.0
C.a
D.None of these
Solution
= a 
(-1)r-1 . nCr - 
r. nCr(-1)r-1 = a[nC1 - nC2 + nC3 .....+(-1)n-1 .nCn] - n 
(-1)r-1 n-1Cr-1
= a(1) - n[n-1C0 - n-1C1 + .......+(-1)(n -1) n-1Cn-1]
= a - n(0) = a
(-1)r-1 . nCr - r. nCr(-1)r-1 = a[nC1 - nC2 + nC3 .....+(-1)n-1 .nCn] - n (-1)r-1 n-1Cr-1= a(1) - n[n-1C0 - n-1C1 + .......+(-1)(n -1) n-1Cn-1]
= a - n(0) = a
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