# Thread: help: Binomial Theorem - Expanding Terms

1. ## help: Binomial Theorem - Expanding Terms

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 +....

----------------------

This is as far as I've gotten and I'm not sure what I should do next:

nay=2y
=> na=12.
=> n=12/a
=> a =12/n

{[n(n-1)]/2} (ay)^2 = 68 y^2.
{[n(n-1)]/2} (a)^2 = 68
{[(n^2) - n]/2} (a)^2 = 68
[(n^2) - n](a)^2 = 136

2. Originally Posted by ilc
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 +....

----------------------

This is as far as I've gotten and I'm not sure what I should do next:

nay=2y
=> na=12.
=> n=12/a
=> a =12/n

{[n(n-1)]/2} (ay)^2 = 68 y^2.
{[n(n-1)]/2} (a)^2 = 68
{[(n^2) - n]/2} (a)^2 = 68
[(n^2) - n](a)^2 = 136
You have 2 unknowns in that.
If you have only one unknown, you can solve.

Hence, use $n=\frac{12}{a}$ to write it in terms of "a" only,

thereby solving for "a".

Or use $a=\frac{12}{n}$ to write it in terms of "n" only,

thereby solving for "n".

Once you've found one, you can find the other.
This is why the problem gave 2 clues.

3. Originally Posted by Archie Meade
You have 2 unknowns in that.
If you have only one unknown, you can solve.

Hence, use $n=\frac{12}{a}$ to write it in terms of "a" only,

thereby solving for "a".

Or use $a=\frac{12}{n}$ to write it in terms of "n" only,

thereby solving for "n".

Once you've found one, you can find the other.
This is why the problem gave 2 clues.

I understand that I have to use a=12/n or n=12/a and substitute it into the equation to solve it but I'm not sure what to do after I substitute it in.

a =12/n
n =12/a

[(n^2) - n](a)^2 = 136
[(n^2) - n](12/n)^2 = 136
what would I do after this?

4. Originally Posted by ilc
I understand that I have to use a=12/n or n=12/a and substitute it into the equation to solve it but I'm not sure what to do after I substitute it in.

a =12/n
n =12/a

[(n^2) - n](a)^2 = 136
[(n^2) - n](12/n)^2 = 136
what would I do after this?
Solve for n.

5. Yes, you must now solve for n.

I would prioritise knowing how to solve the resulting problem,
since expanding the binomial expansion is more complex than that.

understand the following very clearly,

You can also find "a" first...

$an=12\ \Rightarrow\ n=\frac{12}{a}$

$\frac{n(n-1)a^2}{2}=68\ \Rightarrow\ \frac{12}{a}\left(\frac{12}{a}-\frac{a}{a}\right)a^2=136$

$\frac{12}{a}\left(\frac{12-a}{a}\right)a^2=136$

$12(12-a)=136\ \Rightarrow\ 3(12-a)=34\ \Rightarrow\ 12-a=\frac{34}{3}$

$a=12-\frac{34}{3}=\frac{36}{3}-\frac{34}{3}=\frac{2}{3}$

Use this "a" to discover "n".

6. ## completely lost...

Could someone explain how to get even this far. I can't even figure out how to START solving this problem.

"This is as far as I've gotten and I'm not sure what I should do next:

nay=2y
=> na=12.
=> n=12/a
=> a =12/n"

Thanks,

Wyatt...