C2 Binomial Expansion/Theorem Help

In the binomial expansion, in ascending powers of x, of (1+ax)ⁿ, where a and n are constants, the coefficient of x is 15. the coefficient of x² and x³ are equal.

a) Find the value of a and the value of n

I know that I have to expand the brackets only writing down the first 4 terms.
So expanding (1+ax)ⁿ
= 1ⁿ + n[1^(n-1)](ax) + [n(n-1)/2!] [1^(n-2)] [a²x²]
+ {[n(n-1)(n-2)]/3!}1^(n-2)] [1^(n-3)][a³x³]
(Sorry if that looks really confusing)
So simplifying it to make = 1 + nax + (n(n-1)/2) a²x² + (n(n-1)(n-2)/6) a³x³

So the coefficient of x is 15: an = 15 rearranging to make a = 15/n

And the coefficient of x² and x³ are equal:
(n(n-1)/2) a² = (n(n-1)(n-2)/6) a³

It's up to this point where im stuck on..I don't know whether to simplify by cancelling or to expand :confused:

Please help!

Reply 1

Square

This may be horribly horribly wrong, I dont exactly know what to do but I've done something.

Your binomial expansion is the following:

(x+y)^n=\displaystyle \sum_{r=0}^{n} \displaystyle \binom{n}{r}(x)^{n-r}(y)^{r}

You have two unknowns so therefore its best to elimate one by writing it as a form of another. (Simultaneous equations kinda)

The coefficient is going to be this part of the expansion. Since the first term is

1^{n-r}

it is always going to be 1 so it is not going to have an effect on the coefficient of the x terms so it is easier if we just forget about it for now.

[br]\displaystyle \binom{n}{r}(a)^r[br]

We are told it is ascending powers of x, therefore it starts of at r=0 is the number term, r=1 is the x term, r=2 is the x squared term, r=3 is the x cubed term.

Therefore:

\displaystyle \binom{n}{1}a=15[br][br]\displaystyle \binom{n}{2}a^2=\displaystyle \binom{n}{3}a^3[br][br]\displaystyle \binom{n}{2}=\displaystyle \binom{n}{3}a[br][br]\implies a=\displaystyle \binom{n}{2}\div\displaystyle \binom{n}{3}[br][br]

Substitute this into 1:

[br][br]\displaystyle \binom{n}{1}(\displaystyle \binom{n}{2}\div\displaystyle \binom{n}{3})=15

\displaystyle \binom{n}{1}=1 \implies \displaystyle \binom{n}{2}\div\displaystyle \binom{n}{3})=15

n\times(\frac{n!}{2!(n-2)!}\div\frac{n!}{3!(n-3)!})=15[br]\implies n\times\frac{n!}{2!(n-2)!}\times\frac{3!(n-3)!}{n!}=15[br]\implies n\times\frac{3(n-3)!}{(n-2)!}=15[br]

This gives me a quadratic for n=5 or 18. Dont think this is right :/

Reply 2

ukgea

lilydelarocks

And the coefficient of x² and x³ are equal:
(n(n-1)/2) a² = (n(n-1)(n-2)/6) a³

It's up to this point where im stuck on..I don't know whether to simplify by cancelling or to expand :confused:

Please help!

Cancelling is the way to go. Though, you need to be slightly careful not to divide by 0:

First, divide a^2 from both sides of the equation. a^2 cannot be zero, because then a has to be zero which contradicts an = 15.

(n(n-1)/2) = (n(n-1)(n-2)/6) a

Then, divide by n. Again, since an = 15, n cannot be zero.

(n-1)/2 = ((n-1)(n-2)/6)a

Now, we want to divide by n-1. But n - 1 can be zero (n = 1, a = 15 is indeed a solution). What we do then is either

we move stuff around to get an equation where (n - 1) as a factor of something which is to be equal to something that's equal to zero

or

we note that n = 1 is a solution, state clearly that we are now looking for solutions where n is not equal to zero, and then happily divide by n - 1 (which we can do because now we are only concerned with cases where n - 1 not equal to 0)

Going with the latter approach, we are now looking for solutions with n not equal to 0:

1/2 = ((n-2)/6)a

which you should be able to solve.

Reply 3

drizzle

Malcolm Kane

This gives me a quadratic for n=5 or 18. Dont think this is right :/

Yeah when i was doing it my own way by cancelling things out I got n=5 which means a = 3 but when i substituted it back into the formulaes that i made to work out the coefficients of x² and x³ they were different..when they're supposed to be the same.

Reply 4

coffeym

[br](1+ax)^n=1+nax+\frac{n(n-1)a^2x^2}{2}+\frac{n(n-1)(n-2)a^3x^3}{6}+...

Now we can clearly say:

[br]an=15

\frac{n(n-1)a^2}{2}=\frac{n(n-1)(n-2)a^3}{6}

Dealing with the second equation, we see

3=an-2a =>3=15-2a

and so we find

a=6, n=\frac{5}{2}

Of course this only holds for

|x|<\frac16

Reply 5

drizzle

Thanks for the help guys - much appreciated! :smile:

Im stuck on understanding what this question is asking me to do:
Find, in its simplest form, the term independant of x in this expansion (3 marks)

Are they asking me to find the coefficient of x..? or are they asking for something else?