Quick Q on The binomial expansion.

Hi guys, just come across something in the Core Mathematics 2 textbook that I'm a bit stuck on (page 84, q6) <-- just incase I write it out horribly

I've managed to solve this question with a little nudge in the right direction but I was wondering if there was a different, slightly quicker method to solving it. Here's the question;

The coefficient of $x^2$ in the binomial expansion of $(1 + \frac{x}{2})^n$, where n is a positive integer, is 7. Find the value of n.

Okay, so I figured that the coefficient of $x^2$, which is 7, is equal to $\frac{n}{4}$. From that I get 28, and that's where I got stuck. Someone told me that I know the answer is 8 because 28 is in the expansion of $(a+b)^8$, but surely there's a quicker way than sitting there writing out pascal's triangle all the way to 8?!

Any help is greatly appreciated as usual :P

how did you get n/4

Well, I used the binomial expansion as you normally would but with the power as n rather than a number. So the expansion in ascending order of x gave me $\displaystyle \binom{n}{2}$ x $1^n-1$ x $\frac{x^2}{4}$. Then, making 7 = $\frac{n}{4}$ I got 28 as my answer. If I went wrong somewhere, please point it out and please feel free to give me tonnes and tonnes of help

Thanks

You have not used the binomial expansion at all

$(1+\frac{x}{2})^n = 1 + n(\frac{x}{2}) + \frac{n(n-1)}{2!}(\frac{x}{2})^2$

You have not used the binomial expansion at all

$(1+\frac{x}{2})^n = 1 + n(\frac{x}{2}) + \frac{n(n-1)}{2!}(\frac{x}{2})^2$

That's what I tried to type up!

Oh, that's a bit of a problem then . I thought it was possible to use the nCr function to find the coefficients of the terms in the expansion? That's sort of how I got to my answer.. It seems to work, but please explain your method, it seems more expansion-ish. Thanks again lol..

I have used nCr but in the simplified version that you should have also been taught

You seemed to decide that nC2 = 1... why?

7 = $\frac{n}{4}$ I got 28 as my answer.

I didn't intend to make is equal 1. I thought that nC2 was equal to 28, because that divided by 4 would give the coefficient of $x^2$.

It is

Okay, that makes me feel much better, thank you. But now the problem is, I'm not sure how to get from nC2 = 28 to what n is equal to. I figured it was 8 because 28 is in pascal's triangle under the expansion of $(a+b)^8$

Oh god, scrap all of that - I just realised what I was doing wrong. Thank you for your help and patience though!!

nC2 = 28

so, rewrite this using the formula to convert from nCr into factorials:

n! / (n-2)!2! = 28

then you can sort this out by removing the factorials and you should get a quadratic

By using the fact that $^nC_2 = \dfrac{n(n-1)}{2!}$