Coefficient of a term

  • #1

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.
 

Answers and Replies

  • #2
1,425
1
Look at how a product develops as you add more terms, i.e.

[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
1,425
1
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]
 
Last edited:
  • #4
JasonRox
Homework Helper
Gold Member
2,314
3
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.
 

Attachments

  • #5
1,425
1
Hummm... I don't think it's easier for this specific problem. What we are interested in is x^n-1, which coefficient can be found by summing 1 to n and then multiplying by -1. Using combinatorics takes time, as you need to find 3 coefficients and then add them.
 
  • #7
Whoa. Thanks. That's really helpful. Why didnt I see that?!?
 

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