binomial expansion backwards?

the binomial expansion of (1 + ax)^n begins 1 + 2x +3x^2/2

find the value of the constants a and n

how do i go about a question like this?

thanks

Equating coefficients of x and x^2:

na = 2

[n(n-1)/2] a^2 = 3/2

Solve for n and a

*cough* 2! *cough*

ok - heres another part im stuck on.

what is the coefficient of x^n of the function:

1 - 3x +7x -15x .....

how do i go about doing a question like this - i know the answer and see how the answer is true but i dont see how i would have thought it through to get to the answer. Soz if im not making sense.

Two things:

(1) should that be 1 -3x + 7x^2 - 15x^3 +...?

(2) if we don't know more about the series (like it came from (1+ax)^n - is that a given) then the general term could be anything.

yes there should be a squared term there and it comes from

1/(1-x)(1-2x)

when put in partial fractions becomes -1/(1-x) -2/(1-2x)

does that help?

exactly - expand each of them with the binomial expansion and add them up

yes i have done that - to get the binomail expansion shown in my first post about this question ( + the squared term). How do i go about finding the coeffiecient of x^n.

i think i just have to look at the other co efficients and work out the pattern - is there a systematic way to find the pattern in numbers and make a formula for n?

my other bi exp. is more pressing by the way (its on a diff post)- this was just curiosity really.