SOLUTION: The first three terms in the expansion of (1+ay)^n are 1,12y, and 68y^2 Evaluate a and n. Use the fact that: (1+ay)^n=1+nay+n(n-1)/2(ay)^2+... my teacher told me to use:

Question 172778: The first three terms in the expansion of (1+ay)^n are 1,12y, and 68y^2
Evaluate a and n. Use the fact that:
(1+ay)^n=1+nay+n(n-1)/2(ay)^2+...
my teacher told me to use: c(n,k)(1)^n(ay)^k to figure it out and to refer to pascals triangle, But it hasnt helped. I would very much appricate if someone could help.
Thanks
Found 2 solutions by vleith, adamchapman:
Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
First three terms are 1, 12y and 68y^2
Your hint says (1+ay)^n=1+nay+n(n-1)/2(ay)^2+
so (1+ay)^n=1+12y+68y^2+...
12y=nay, so na=12 and a=12/n
68y^2 = (n(n-1)/2)(ay)^2 = (n(n-1)/2)a^2y^2. so (n(n-1)/2)a^2 = 68
Now substitute the value for a into the second equation and solve

so a = 2/3
now check against the hint to see if the values work
a*n = (2/3)*18 = 12. so far so good
(18*17/2)*(4/9) = 68

Answer by adamchapman(301) (Show Source): You can put this solution on YOUR website!
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

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:

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:

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)
...(2)
...(3)
adding (2) and (3) together, we get:

Rearranging to get zero on the right hand side, we have:

which can be solved with

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
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