# Homework Help: Finding a pattern for a series and the general formula

1. Feb 11, 2013

### PhizKid

1. The problem statement, all variables and given/known data
Given:

When n = 1:
$-1 + xln(e)$

When n = 2:
$2 - 2xln(e) + x^{2}ln(e)^{2}$

When n = 5:
$-120 + 120xln(e) - 60x^{2}ln(e)^{2} + 20x^{3}ln(e)^{3} - 5x^{4}ln(e)^{4} + x^{5}ln(e)^{5}$

When n = 7:
$-5040 + 5040xln(e) - 2520x^{2}ln(e)^{2} + 840x^{3}ln(e)^{3} - 210x^{4}ln(e)^{4} + 42x^{5}ln(e)^{5} - 7x^{6}ln(e)^{6} + x^{7}ln(e)^{7}$

Identify the series and write a general formula for it (in sigma summation notation if possible).

2. Relevant equations
N/A

3. The attempt at a solution

The pattern I do see is that the constant is always n!, but when n = 2, this factorial is not negative because there are an even number of terms, and the signs alternate starting from the highest power n where this term is positive.

So the highest nth term does not have any factorials, but the subsequent terms, alternating signs, begin to increase in factorials based on what n is. For example, when n = 5, the next highest term has coefficient 5, the next term has 5*4, then 5*4*3, and finally 5*4*3*2.

I'm not sure how to deal with the alternating signs in sigma summation notation or the factorials part.

2. Feb 12, 2013

### jbunniii

This might help with a formula for the coefficients:
5*4*3*2*1 = (5!)/(0!)
5*4*3*2 = (5!)/(1!) (which is also equal to (5!)/(0!))
5*4*3 = (5!)/(2!)
5*4 = (5!)/(3!)
5 = (5!)/(4!)
1 = (5!)/(5!)

Also, (-1)^n = 1 if n is even, -1 if n is odd.

So try writing something of the form
$$\sum_{k = 0}^{n}(\textrm{formula involving factorials})(\textrm{formula involving }(-1)^k)(\textrm{formula involving power of }x \ln(e))$$