Question 172778
Pascal's triangle is just a list of of binomial expansion terms. Check it out at http://library.thinkquest.org/C0110248/algebra/biexpintro.htm .

The
{{{c(n,k)(1)^n(ay)^k}}}
is the quation used to work out the kth component of the nth binomial expansion term.
e.g. the first part (k=1) of the n=3 term would be:
{{{c(3,1)(1)^3(ay)^1=ay}}}
It is simply an individual number on pascals triangle. Dont bother trying to obtain a deep understanding straight away, look at things ion the simplest level and try to reproduce them.  You could try expanding (1+1) using that equation and reproduce pascal's triangle.



Since the first term in your series is equal to one, I would assume that the first term is what some consider as the zeroth term (i.e. (1+ay)^0)

Absolutely anything to the power of zero is one. Your teacher might have proved it to you before, but heres a quick proof:

{{{x^2/x=x^1=x}}}
{{{x^2/(x^2)=1=x^0}}}

anyway, let's look back at the binomial expansion. Replacing "x" with "ay" in pascals triangle at the link i gave, and setting the right hand side of the following expressions to the values given in your question:

{{{(1+ay)^0=1}}}                         ...(1)
{{{(1+ay)^1=1+ay=12y}}}                  ...(2)
{{{(1+ay)^2=1+2ay+(ay)^2=68y^2}}}        ...(3)

adding (2) and (3) together, we get:
{{{(1+ay)+(1+2ay+(ay)^2)=12y+68y^2}}}
{{{(1+2ay+(ay)^2)+ay(1+2ay+(ay)^2)=12y+68y^2}}}
{{{1+2ay+a^2 y^2+ay+2a^2 y^2+a^2 y^2=12y+68y^2}}}
{{{1+3ay+3 a^2 y^2=12y+68y^2}}}

Rearranging to get zero on the right hand side, we have:
{{{(3a^2-68)y^2 + (3a-12)y + 1 =0}}}

which can be solved with
{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


I'll let you try that to find the value of "a" which satisfies the quadratic.


Don't bother trying to work out what the value of n is you have already used n=0,1,2 to work out the first three terms of the binomial expansion. 

I hope this helps
Adam