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# C2 Binomial Expansion Question 3 watch

1. The first three terms in the expansion of (a + 3x)^n are 16b, 144bx and 540bx^2.
Find the values of the constants a, b and n.

I'm pretty much clueless, and really need help! Thanks.
2. Well, what are the first 3 terms in the expansion of (a+3x)^n? Use the binomial theorem.
3. DFranklin strikes again with the most useless of posts.
4. I can get 'a' but Im stuck after that!
5. (Original post by Fraser_Integration)
DFranklin strikes again with the most useless of posts.
Lolwut? As far as I can tell his post was actually quite useful if you followed it.
6. (Original post by Fraser_Integration)
{mod: abusive comment deleted}

Fraser_Integration strikes again with the most useless of posts
7. (Original post by DFranklin)
Well, what are the first 3 terms in the expansion of (a+3x)^n? Use the binomial theorem.
That should give you 3 simultaneous equations.

These sorts of questions always end up in simultaneous equations.
8. (Original post by stevencarrwork)
That should give you 3 simultaneous equations.
Sort of. They won't be terribly 'simultaneous' though. (i.e. you can easily solve them by substitution).

These sorts of questions always end up in simultaneous equations.
Not any more. Not that standards have fallen since our day, of course...
9. (Original post by DFranklin)
Not any more. Not that standards have fallen since our day, of course...
I always found those questions really dull. It was just processing symbols, with no thinking needed.

You needed to memorise the binomial expansion, but once you did, the questions were so trivial.

The questions always came down to '7b is 4 times (7*6/2))*b^2 What is b?'

The questions were also never useful in other topics.

How often do you have a binomial expansion where you know the second coefficient is 3 times the first, and the third is 8 times the second, and you have to work out what the power was?
10. This question appears on p.285 of OUP's Edexcel Core C1 C2 book. It's considerably harder than you'll meet in an actual exam, so don't be worried if you can't do it. Here's the solution for anyone still looking.

First expand (a+3x)^n to the first three terms, applying the following binomial expansion formula:

(a + bx)^n = (nC0)a^n + (nC1)a^(n-1)bx + (nC2)a^(n-2)(bx)^2 + ...

If we do this correctly, we end up with the following three terms:

(1) a^n
(2) n * a^(n-1) * 3x = 3na^(n-1)x
(3) n(n-1)/2! * a^(n-2) * (3x)^2 = 9/2(n)(n-1)a^(n-2)x^2

We're told that these three terms correspond to 16b, 144bx, and 540bx^2 respectively. From this we obtain the following three simultaneous equations:

(1) 16b = a^n
(2) 144b = 3na^(n-1)
(3) 540bx^2 = 9/2(n)(n-1)a^(n-2)

So now we have to solve these for a, b, and n. But it's a little tricky, so I'm going to step through this part as well. Here goes. Looking at the first two equations, we see that 9 * 16b = 144b. So,

9 * a^n = 3na^(n-1)
3a^n = na^(n-1) [dividing both sides by 3]
3/n = a^(n-1) / a^n [dividing both sides by n and a^n]
3/n = a^-1 = 1/a
[subtracting indices for division]
a = n/3 [rearranging]
n = 3a [rearranging]

Looking at the last two equations, we see that 15/4 * 144b = 540b. So,

15/4 * 3na^(n-1) = 9/2(n)(n-1)a^(n-2)
5/2 * a^(n-1) = (n-1)a^(n-2) [dividing both sides by 9n/2]
a^(n-1) / a^(n-2) = 2/5(n-1) [dividing both sides by 5/2 & a^(n-2)]
a = 2/5(n-1) [subtracting indices for division]

We now have two definitions of a: a = n/3 and a = 2/5(n-1). Therefore:

n/3 = 2/5(n-1) [since a = a]
5n = 6(n-1) = 6n-6 [multiplying both sides by 3*5]

Now at last we're there!

n = 6
[simplifying above]
a = 2 [since a = n/3]
b = 4 [since 16b = a^n]
11. (Original post by Timotheos)
Now at last we're there!
At last is an understatement! 3 years and 22 days since the last post

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