Question

How do I find the coefficient of y^99 in the expansion
(y-1)(y-2)(y-3)...........(y-100)

Please I need details solution

Answer 1

1 , 2 , 3...100 are zeroes of the polynomial.

Let P(n) = ( y - 1 ) ( y - 2 )...( y - n )

P(1) = y - 1

P(2) = ( y - 1 ) ( y - 2 ) = y^2 - ( 1 + 2 ) ∙ y + 1 ∙ 2

P(3) = ( y - 1 ) ( y - 2 ) ( y - 3 ) = y^3 - ( 1 + 2 + 3 ) ∙ y^2 + 11 y - 1 ∙ 2 ∙ 3

P(n) = y^n - ( 1 + 2 + 3 +…+ n ) ∙ y^( n - 1 ) + … + ( ( - 1 ) ^n ) ∙ 1 ∙ 2 ∙ 3 ∙ n

So:

P(100) = y^100 - ( 1 + 2 + 3 +…+ 100) ∙ y^99 + … + 1 ∙ 2 ∙ 3…∙ 100

The coefficient of y^99 = - ( 1 + 2 + 3 + … + 100 )

The sum of first n natural numbers, 1 + 2 + 3 +… + n = ( n + 1 ) ∙ n / 2

The coefficient of y^99 = - ( 1 + 2 + 3 + … + 100 ) = - ( 100 + 1 ) ∙ 100 / 2 = - 101 ∙ 50 = - 5050