Polynomial of degree 100

I noted the same thing as others, but started with multiplying all numbers from 1 to 100 by every other number, so 5050*5050. But this also counts the squares, so need to subtract the sum of n^2 from n=1-100. n(n-1)(2n-1)/6. Now we have counted everything twice, so we just need to divide by 2. (25502500-338350)/2

The Polynomial can be written as :

$(x-1)(x-2)(x-3).........................................(x-100)$

For getting the coefficient of $x^{98}$ we need to multiply $x$ from 98 terms and the constants from the remaining 2 terms.

By this the coefficient $x^{98}$ will come out to be :

$1(2+3+4+........+100)$ $\ + \ 2(3+4+5+.......+100)$ $\ + \ 3(4+5+6+........+100)$ $\ + \ .....and \ so \ on$

This can be represented as a series with : $T_{n}=n(5050-\sum n)$

Note that 5050 is the sum of first 100 Natural no.s or sum of all the roots of polynomials.

We need to Find :

$\sum T_{n}=\dfrac{5050(n)(n+1)}{2}-\dfrac{[n(n+1)]^{2}}{8}-\dfrac{n(n+1)(2n+1)}{12}$

Put $n=100$ to get $S_{n}=12582075$

So coefficient of $x^{98}$ is 12582075

use the identity( a1+a2+a3+a4+.....+an)^2=a1^2+....an^2+2(product of two roots taken two at a time )

To find

$1(2 + 3 + 4 + ... + 100) + 2(3 + 4 + ... + 100) + ... + 98(99 + 100) + 99(100),$

we can find the general term for the sequence

$(2 + 3 + 4 + ... + 100), (3 + 4 + ... + 100), ... , (99 + 100) , 100$

or

$5049, 5047, 5044, ... , 199, 100.$

By the method of differences, we see that the general term is a quadratic one. We therefore pick three ordered pairs to solve for such equation, say $(1,5049), (2, 5047), (3, 5044)$ . Eventually, we can solve for the general term as $0.5(-k^2 - k + 10100)$ .

The original sequence then can be represented as

$1(2 + 3 + 4 + ... + 100)+ 2(3 + 4 + ... + 100) + ... + k(0.5(-k^2 - k + 10100)) + ... 99(100)$

or more compactly as

$\displaystyle \sum_{k = 1}^{99}$ $0.5(-k^3 - k^2 + 10100k) = 12582075$

Let $S_n = \displaystyle\sum_{i=1}^{100}i^n$

By Vieta's, $-a_{99} = S_1 = 5050$

Using Newton's Sums:

$S_2+a_{99}S_1+2a_{98} = 0$

$\dfrac{100 \times 101 \times 201}{6}-5050^2+2a_{98} = 0$

$a_{98} = \dfrac{25502500-338350}{2} = \boxed{12582075}$