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:

Algebra ->  Permutations -> 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:       Log On


   

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) About Me  (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
n%28n-1%29a%5E2%2F2+=+68
%28%28n%5E2-n%29%2F2%29+%2A+%2812%2Fn%29%5E2+=+68
%28n%5E2-n%29%28144%2F2%29+%2F+%28n%5E2%29+=+68+
%28n%5E2+-n%2972+=+68n%5E2
72n%5E2+-+72n+=+68n%5E2
4n%5E2+=+72n
4n+=+72
n+=+18
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) About Me  (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
c%28n%2Ck%29%281%29%5En%28ay%29%5Ek
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%283%2C1%29%281%29%5E3%28ay%29%5E1=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%5E2%2Fx=x%5E1=x
x%5E2%2F%28x%5E2%29=1=x%5E0
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:
%281%2Bay%29%5E0=1 ...(1)
%281%2Bay%29%5E1=1%2Bay=12y ...(2)
%281%2Bay%29%5E2=1%2B2ay%2B%28ay%29%5E2=68y%5E2 ...(3)
adding (2) and (3) together, we get:
%281%2Bay%29%2B%281%2B2ay%2B%28ay%29%5E2%29=12y%2B68y%5E2
%281%2B2ay%2B%28ay%29%5E2%29%2Bay%281%2B2ay%2B%28ay%29%5E2%29=12y%2B68y%5E2
1%2B2ay%2Ba%5E2+y%5E2%2Bay%2B2a%5E2+y%5E2%2Ba%5E2+y%5E2=12y%2B68y%5E2
1%2B3ay%2B3+a%5E2+y%5E2=12y%2B68y%5E2
Rearranging to get zero on the right hand side, we have:
%283a%5E2-68%29y%5E2+%2B+%283a-12%29y+%2B+1+=0
which can be solved with
y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

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