Binomial Expansion (Core 2, AQA)

Struggling with the following two questions, so if anyone could help that would be appreciated. Found most questions on binomial expansion relatively easy but I can't think of how to work out this question, as far as I know I haven't been taught how to, nor is it in the textbook, perhaps it relies on intuition (which I then clearly don't possess).

Anyway;
15. In the binomial expansion of(1+px)ⁿ, where p and n are constants and n is a positive integer, the coefficient of x is -12 and the coefficient of x² is 60. Find the value of p and the value of n.

16. In the expansion of (1-x)(1+2x)ⁿ the coefficient of x² is 198. Find the value of n and the coefficient of x³ in the expansion.

Thanks!

Struggling with the following two questions, so if anyone could help that would be appreciated. Found most questions on binomial expansion relatively easy but I can't think of how to work out this question, as far as I know I haven't been taught how to, nor is it in the textbook, perhaps it relies on intuition (which I then clearly don't possess).

Anyway;
15. In the binomial expansion of(1+px)ⁿ, where p and n are constants and n is a positive integer, the coefficient of x is -12 and the coefficient of x² is 60. Find the value of p and the value of n.

16. In the expansion of (1-x)(1+2x)ⁿ the coefficient of x² is 198. Find the value of n and the coefficient of x³ in the expansion.

Thanks!

Okay sorry guys, if you would be able to do so, some further prodding would be greatly appreciated; after considering the problem with your help I still can't think of how to answer either question.

Let us try question 15 and see if we can see a patern in the various values that n can take:

Please do not give full solution

No full solution was supplied nevertheless. I left it up to the reader to deduce how it is known that the (px)^2 coefficient is given by n(n-1)/2.

Let us try question 15 and see if we can see a patern in the various values that n can take:

For n = 1, (1+px)^1 = 1 + px [coef of x is p]

For n = 2, (1+px)^2 = 1 + 2px + (px)^2 [coef of x is 2p]

For n = 3, (1+px)^3 = 1 + 3px + 3(px)^2 + (px)^3 [coef of x is 3p]

For n = 4, (1+px)^4 = 1 + 4px + 6(px)^2 + 4(px)^3 + (px)^4

Notice that the coefficient of x is given by np. For example, when n = 3, the coefficient of x is 3p etc.

So if the coefficient of x is -12, then 3p = -12, and therefore p = -4.

Thus we have (1 - 4x)^n, and we still need the value of n.

The coefficient of x^2 is given by pn(n-1)/2 = -4n(n-1)/2. Therefore set this to -60 to find n.

-60 = -4n(n-1)/2
30 = n^2 - n
n = 6

Therefore the original expression is (1-4x)^6

What did you get when you expanded for question 15

Just so you know, I haven't read the reply which gives me the full answer - though my full appreciation goes out to him for his help; essentially, though, I'm where I was before he replied.

So, I did the typical thing I do with binomials and got n choose 1 multiplied by px, and then n choose 2 multiplied by (p)^2(x)^2.

What do you understand n choose 2 to be?

Erm well, I know it's related to Pascal's Triangle, rather embarrassingly it's just something I type in on my calculator when it is two known numbers. I also know of a formula, n!/k!(n-k)! = n choose k, but I'm not sure if that's related.

Ok

Try some values
3 choose 1
4 choose 1
5 choose 1

Etc

Can you see what n choose 1 is

Well it would seem as though it would be n. I think I noticed this before but not sure what to do with it

P is not -4. You have used n=3 to deduce that but then go on to say n=6

Ok so now you have

1 + npx + (n choose 2)(px)^2

So you know that np = -12

Now

Consider
3 choose 2
4 choose 2
5 choose 2

Look for a pattern there

If you are struggling try

Okay, well the pattern I see is that when you increase n by one for choose 2 the increase in the total goes up by one each time, based upon the previous increase (if that makes sense)

Also, (n-1) choose 1 + (n-1) choose 2 seems to equal n choose 2

Ok

nC2 is n(n-1)/2

This is because, as you knew, nC2 = n!/2!(n-2)!

The n! And the (n-2)! Cancel leaving n(n-1) in the numerator

Sorry, would you be able to offer me some further help this particular point, I'm really quite stuck - nothing I've done in AS level seems to be of this difficulty! Phew. Thanks.