- #1

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## Homework Statement

What is the coeff of [tex]x^{99}[/tex] in (x-1)(x-2)...(x-100)

**2. The attempt at a solution**

This has to do with the binomial coeff. I dont know how to go about it.

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- Thread starter chaoseverlasting
- Start date

- #1

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What is the coeff of [tex]x^{99}[/tex] in (x-1)(x-2)...(x-100)

This has to do with the binomial coeff. I dont know how to go about it.

- #2

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[tex](x-1)(x-2) = x^2 - (1+2)x + 2 [/tex]

[tex](x-1)(x-2)(x-3) = x^3 -(1+2+3)x^2 + 11x + 6[/tex]

now, what do you see?

- #3

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Look at how a product develops as you add more terms,

if

[tex](x-1)(x-2)...(x-n) = x^n - (1+2+...+n)x^{n-1} + ... + (-1)^{n}*1*2*...*n [/tex]

then

[tex](x-1)(x-2)...(x-n)(x-(n+1)) = x^{n+1} - (1+2+....+n+n+1)x^{n} + ... + (-1)^{n+1}*1*2*...*n*(n+1) [/tex]

If we let n + 1 = m, then

[tex](x-1)(x-2)...(x-m) = x^m - (1+2+...+m)x^{m-1} + ... + (-1)^{m}*1*2*...*m [/tex]

if

[tex](x-1)(x-2)...(x-n) = x^n - (1+2+...+n)x^{n-1} + ... + (-1)^{n}*1*2*...*n [/tex]

then

[tex](x-1)(x-2)...(x-n)(x-(n+1)) = x^{n+1} - (1+2+....+n+n+1)x^{n} + ... + (-1)^{n+1}*1*2*...*n*(n+1) [/tex]

If we let n + 1 = m, then

[tex](x-1)(x-2)...(x-m) = x^m - (1+2+...+m)x^{m-1} + ... + (-1)^{m}*1*2*...*m [/tex]

Last edited:

- #4

JasonRox

Homework Helper

Gold Member

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Look at how a product develops as you add more terms,

if

[tex](x-1)(x-2)...(x-n) = x^n - (1+2+...+n)x^{n-1} + ... + (-1)^{n}*1*2*...*n [/tex]

then

[tex](x-1)(x-2)...(x-n)(x-(n+1)) = x^{n+1} - (1+2+....+n+n+1)x^{n} + ... + (-1)^{n+1}*1*2*...*n*(n+1) [/tex]

If we let n + 1 = m, then

[tex](x-1)(x-2)...(x-m) = x^m - (1+2+...+m)x^{m-1} + ... + (-1)^{m}*1*2*...*m [/tex]

There is an easier way.

Here, I'll post a picture of some of hour lecture notes.

If you understand this example, any other question like it will be a breeze.

- #5

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- #6

- 682

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Vieta's formula

http://mathworld.wolfram.com/VietasFormulas.html

http://mathworld.wolfram.com/VietasFormulas.html

- #7

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Whoa. Thanks. That's really helpful. Why didnt I see that?!?

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