Question

what is the coefficient of x^99 in polynomial (x-1)(x-2)......(x-100)?​

Answer 1

Just consider the binomial  x - 1.  Here the coefficient of x⁰ is - 1.

Consider  $(x-1)(x-2)=x^2-3x+2.$  Here the coefficient of x is  - 1 - 2 = - 3.

Consider  $(x-1)(x-2)(x-3)=x^3-6x^2+11x-6.$  Here the coefficient of x² is  - 1 - 2 - 3 = - 6.

So, is the given below true?!

$\boxed{\text{Coefficient of \ x^{n-1} \ in the expansion of \ \displaystyle\prod_{i=1}^n(x-i) \ is \ -\dfrac{n(n+1)}{2}. }}$

Let's prove it by principle of mathematical induction!

We found P(1), P(2) and P(3) are true. So we assume P(k) as true.

$\text{Let the coefficient of \ x^{k-1} \ in the expansion of \ \displaystyle\prod_{i=1}^k(x-i) \ be \ -\dfrac{k(k+1)}{2} .}$

Let  $\displaystyle\prod_{i=1}^k(x-i)\ =\ x^k-\dfrac{k(k+1)}{2}x^{k-1}+r,$  where  $r$  represents the rest of the expansion.

Considering P(k+1)...

$\displaystyle\prod_{i=1}^{k+1}(x-i)\ =\ (x-1)(x-2)(x-3)\dots\dots(x-k)(x-k-1)\\ \\ \\ =\ \left(x^k-\dfrac{k(k+1)}{2}x^{k-1}+r\right)(x-k-1)\\ \\ \\ =\ (x^k+r)(x-k-1)-\dfrac{k(k+1)}{2}x^{k-1}(x-k-1)\\ \\ \\ =\ x^{k+1}-kx^k-x^k+rx-rk-r-\dfrac{k(k+1)}{2}x^k+\dfrac{k^2(k+1)}{2}+\dfrac{k(k+1)}{2}\\ \\ \\ =\ x^{k+1}-\left(k+1+\dfrac{k(k+1)}{2}\right)x^k+(x-k-1)r+\left(\dfrac{k(k+1)}{2}\right)(k+1)$

Taking the coefficient of  $x^k,$     [Since n = k + 1]

$-\left(k+1+\dfrac{k(k+1)}{2}\right)\ =\ -\dfrac{2(k+1)+k(k+1)}{2}\\ \\ \\ =\ -\dfrac{(k+1)(k+2)}{2}\ =\ -\dfrac{n(n+1)}{2}$

So we proved it! It's true!

Thus we can say that,

$\text{Coefficient of \ x^{99} \ in \ \displaystyle\prod_{i=1}^{100}(x-i)\ =\ -\dfrac{100\cdot 101}{2}\ =\ }\mathbf{-5050}$

Hence -5050 is the answer.