# Binomial expansion

• Aug 12th 2009, 05:07 AM
jake.davis
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?
• Aug 12th 2009, 05:31 AM
Hello jake.davis
Quote:

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 $y$ and $y^2$, we have to find the values of $a$ and $n$, for which:

$12 = na$

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

Square the first equation:

$144 = n^2a^2$

If we divide the second equation by this one:

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

Can you continue?

• Aug 12th 2009, 05:46 AM
jake.davis
Upon continuing I determined a=0, n=12. Does that seem right?
• Aug 12th 2009, 06:08 AM
Quote:

Originally Posted by jake.davis
Upon continuing I determined a=0, n=12. Does that seem right?

Starting with equation I gave you:

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

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

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

$\Rightarrow n = 18$

$\Rightarrow 18a = 12$

$\Rightarrow a = \tfrac23$