Question

In expansion of (x-1)(x-2)....(x-100) what will be the coefficient of x^99 ?

Answer 1

Given: f(x) = (x - 1)(x - 2)(x - 3) ........... (x - 100)

ii) Expanding the above, we get:
f(x) = x¹°° + a*x⁹⁹ + bx⁹⁸ + c*x⁹⁷ + ......... + (-1)*(-2)*(-3)...*(-100)
So this f(x) is a polynomial in x¹°°.
Solving this to zero, roots of this polynomial are: {1, 2, 3, ... , 100}

Hence by theory of equations,
Sum of roots = -coefficient of x⁹⁹/coefficient of x¹°° =

In this case, -coefficient of x⁹⁹/1 = Sum of roots

coefficient of x⁹⁹ = -(Sum of roots) = -(1 + 2 + 3 + ... + 100)

[1 + 2 + 3 + ... + 100 is the sum of first 100 natural numbers
Applying sum of 1st n natural numbers = n(n + 1)/2, this sum = 100*101/2]

coefficient of x⁹⁹ = -100*101/2 = -5050

Or,


Given polynomial:

(x – 1) (x – 2) (x – 3) . . . . . . (x – 100)

= x100 – (1 + 2 + 3 + . . . . . . . + 100) x99 + (. . .) x98 + . . . . . .

Here coeff. Of x99 = -(1 + 2 + 3 + . . . . . . .+ 100)

= -100 x 101/2 = -5050.