Question

There's this polynomial in \(x\) with degree 100. The roots of this polynomial are 1, 2, 3... 100. What is the coefficient of \(x^{98}\)?

Answer 1

x^100 - ( 1 + 2 + 3 + ... + 100 ) * x^99 + (1*2 + 1*3 + 1*4 + ... ) * x^98

Answer 2

The polynomial is monic - forgot to mention that.

And yes, that's what I used.

Answer 3

Hold on, please let me type my attempt. :|

Answer 4

\[1 \times (2 + 3 + \cdots + 100) + \]\[2 \times (3 + 4 + \cdots + 100) +\]\[\vdots \]\[99 \times 100\]

Answer 5

Then I used sigma notation.

Answer 6

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bi%3D1%7D%5E%7B100%7D%5Csum%5Climits_%7Bj%3D1%7D%5E%7B100%7D+%28i*j%29

Answer 7

\[\sum n \times \left[\frac{100 - n}{2} \left(2 (n+1) + n - 1\right)\right]\]

Answer 8

^ Am I right till this step? Because it ends up as a very complicated expression.

After seeing @ganeshie8's reply though, I think I am supposed to get a complicated expression.

Answer 9

Wait, I'm not...

Answer 10

\[\sum n \times \left[\dfrac{100-n}{2} \left(2(n+1) + 99 - n\right)\right]\]

Answer 11

Sounds better.\[\sum n \times \dfrac{100- n}{2}\times (n+1)\]

Answer 12

Derp.

Answer 13

can i try ?

Answer 14

\[\sum n\times \dfrac{100 - n}{2}\times (n+101)\]Now we'll just use the results for \(\sum n\), \(\sum n^2\), \(\sum n^3\). But because I'm too lazy to do that, I'd let @perl do the rest of the blabber. :P

Answer 15

well using your idea

Answer 16

Makes sense.