Do you know how to multiply?

Since I'm lacking time I'll straightaway copy $my$ solution from here

Given is a product of 100 factors out of which we have to select 98 x's and 2 constants to get $x^{98}$ i.e $1(2+3..+100)+2(1+3+4...+100)+..+100(1+2+..+99)$ $=\displaystyle \sum_{i=1}^n i(\small{\displaystyle \sum_{r=1}^i (r)}-i)$ $=\displaystyle \sum_{i=1}^n i\dfrac{(i(i-1))}{2}=\dfrac{1}{2}\displaystyle \sum_{i=1}^n (i^3-i^2)$ $\small( {\color{#20A900}{Using~\displaystyle \sum_{i=1}^n i=\dfrac{(n)(n+1)}{2}\\~,~\displaystyle \sum_{i=1}^n i^2=\dfrac{n(n+1)(2n+1)}{6}~and~\displaystyle \sum_{i=1}^n i^3=\dfrac{n^2(n+1)^2}{4}})}$ $=\dfrac{n(n+1)(n-1)(3n+2)}{24}$ Putting n=100, we get the desired result i.e $\Large 12582075$

Let $P(x)=\displaystyle \sum^{100}_{n=1}(x+n)$ and $P_{n}=\sum x^n$ .

So, clearly $P_{1}=-\frac{100\times 101}{2}=-5050$ and $P_2=\frac{100 \times 101 \times 201}{6}=338350$ .

$\therefore P(x)=x^{100}+5050x^{99}+mx^{98}+\ldots$

Now, by Newton's Sums,

$P_2+5050P_1+2m=0 \\ 2m=5050^2-338350=25164150 \\ \Rightarrow m=\boxed{12582075}$