1. ## Binomial expansion

I've been given the first three terms in the expansion of (1+ay)^n:

1, 12y, and 68y^2.

I'm asked to evaluate a and n using the fact that:
(1+ay)^n = 1 + nay + n(n-1)/2(ay)^2 +.....

Any help in this problem would be appreciated, what is meant by evaluate?

2. Hello jake.davis
Originally Posted by jake.davis
I've been given the first three terms in the expansion of (1+ay)^n:

1, 12y, and 68y^2.

I'm asked to evaluate a and n using the fact that:
(1+ay)^n = 1 + nay + n(n-1)/2(ay)^2 +.....

Any help in this problem would be appreciated, what is meant by evaluate?
Evaluate means 'find the value of ...'

So, by comparing the coefficients of $\displaystyle y$ and $\displaystyle y^2$, we have to find the values of $\displaystyle a$ and $\displaystyle n$, for which:

$\displaystyle 12 = na$

and $\displaystyle 68 = \frac{n(n-1)}{2}\cdot a^2$

Square the first equation:

$\displaystyle 144 = n^2a^2$

If we divide the second equation by this one:

$\displaystyle \frac{68}{144}= \frac{n-1}{2n}$

Can you continue?

3. Upon continuing I determined a=0, n=12. Does that seem right?

4. Originally Posted by jake.davis
Upon continuing I determined a=0, n=12. Does that seem right?
Starting with equation I gave you:

$\displaystyle \frac{68}{144}=\frac{n-1}{2n}$

$\displaystyle \Rightarrow \frac{17}{36}=\frac{n-1}{2n}$

$\displaystyle \Rightarrow 34n = 36(n-1) = 36n - 36$

$\displaystyle \Rightarrow n = 18$

$\displaystyle \Rightarrow 18a = 12$

$\displaystyle \Rightarrow a = \tfrac23$