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If (1+x)^n = C0 +C1x + C2 (x)^2.......+Cn(x)^n , then C0Cn – C1Cn-1 + C2Cn-2 – …...+ (-1)^nCnC0 = ?? I tried to solve it using (1+x)^n = C0 +C1x + C2 (x)^2.......+Cn(x)^n …..1 (1 – 1/x)^n = C0 – C1/x + C2 /(x)^2.......+ (-1)^n Cn/(x)^n ...2 Then multiplying 1 and 2 and comparing the coefficients of x^-n of both sides.But I did not get the answer.Could you please suggest where I am wrong??

2 years ago

Answers : (1)

Aditya Gupta
2065 Points
							
(1+x)^n =  C0 +C1x + C2 (x)^2.......+Cn(x)^n  
and (x-1)^n =  Cnx^n – Cn-1x^(n-1) + Cn-2 (x)^n-2 –.......+(-1)^nC0(x)^0
so, C0Cn – C1Cn-1 + C2Cn-2 – …...+ (-1)^nCnC0 = coefficient of x^n in (1+x)^n*(x-1)^n 
=coefficient of x^n in (x^2-1)^n, whose general term is nCk*x^2(n-k)*(-1)^k
so, 2(n-k)=n or n=2k or k=n/2
so, if n is odd, C0Cn – C1Cn-1 + C2Cn-2 – …...+ (-1)^nCnC0= 0
if n is even, C0Cn – C1Cn-1 + C2Cn-2 – …...+ (-1)^nCnC0=  nC(n/2)*(-1)^(n/2)
2 years ago
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