Question

What is the coefficient on x^(n-1) on the expansion of (x+1)(x+2)(x+3)...(x+n) ?

Answer 1

you want a guess?

Answer 2

my guess is \(n!\) but i could easily be wrong
experiment and see what it looks like for \(n=3,4,5\)

Answer 3

Actually n! is the coefficient on the x^0 term =)

Answer 4

yeah my answer was way off

Answer 5

the sequence is 1,3, 6, 10,.....

Answer 6

is the next one 15?

Answer 7

I mean: (x+1)(x+2) , the coefficient of x^1 = 1
(x+1)(x+2)(x+3) , the coefficient of x^2 is 3

Answer 8

and so on...

Answer 9

yes it is , and then 21

Answer 10

sequence 1,3,6,10,15, 21 would mean you have a quadratic

Answer 11

you can tell it is quadratic because the differences go up by one each time

Answer 12

i could be wrong, but it might be \(\frac{1}{2}(n^2+n)\) try that

Answer 13

better guess than \(n!\) for sure!

Answer 14

Yeah that's correct =) Now I dare someone to try to find the x^(n-2) or x^(1) coefficients, or the general formula for any coefficient. =)

Answer 15

oh, i didn't realize this was a test

Answer 16

Oh yeah just a fun game I'm playing haha

Answer 17

do you know the answer?

Answer 18

and it is it a quadratic as well?

Answer 19

i meant "polynomial" not quadratic

Answer 20

Nope, it's not quadratic as far as I know. For instance, the x^1 term's coefficient is \[\Large n \sum_{k=1}^n \frac{1}{k}\] But I don't know the closed form of this, if there even is one. The weird part is I know the form of all the higher terms but I have no idea how to even express them with summation signs.

Answer 21

What do you know about Stirling numbers?

Answer 22

http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Table_of_values_for_small_n_and_k

Answer 23

Huh, no I have never heard of these before, but this is quite interesting.

Answer 24

Here is my sort of insight that I don't know how to write: \[\Large f(x) = (x+a)(x+b)(x+c) \\ f(x) = abc[ (\frac{1}{abc})x^3 + (\frac{1}{ab}+ \frac{1}{ac}+\frac{1}{bc})x^2+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})x + 1]\] So each coefficient on the x^k term is really just a sum of fractions of all the possible ways to pick k of the roots and multiply them together.