Question

In the expansion of (1+3x)^n, the coefficient of x^3 is 6 times the coefficient of x^2.

Show that nC3 = 2(nC2)

*this topic is under binomial theorem

I can't seem to figure this out. How do you know how many multiples something with a choose is compared to the other if there's an unknown??

Answer 1

I don't believe this is true for all n (just take n=7, although it does work for n=8). Is the question asking for which values of n they are equal?

Answer 2

No, I think they're asking to show that the ratio of nC2 to nC3 is 2:1.

There isn't a need to find what n is because thats part 2 of the question.

Answer 3

Well it doesnt seem to work for all n because.. take n =7 for example

\[ 7C3=35\\
2(7C2)=2(21)=42\\
35 \ne 42\]

Answer 4

i'm sorry, are you trying to say that the question is unsolvable?

Answer 5

Do they not give you a specific value of n? Because just expanding \((1+3x)^n\), for any \(n\), and looking at the \(x^2\) and \(x^3\) terms gives:

\[ \frac{9}{2}(n-1)nx^2\] and \[ \frac{9}{2}(n-2)(n-1)nx^3\]. looking at the coefficients, we see that the x^3 terms is (n-2) times bigger. If they claim it is 6 times bigger, then \((n-2)=6 \implies n=8\). And yes the formula works for n=8, but not all n.

Answer 6

nah they don't, that's all i was given

i think my problem was at the middle part where you expanded the x^3 coefficients
how did you do it?
because i expanded it like this and it seems to be wrong somehow



i cant get your answer