Question

I'm not so sure how to do this,
http://i.imgur.com/g1a57bi.png
I know it has something to do with binomial.

Answer 1

The binomial theorem:
\[\sum^n_{k=0} \binom{n}{k} x^{n-k} y^k\]
Substitute x = x, and \(\frac{1}{x^9}\) for y.

Answer 2

I know the binomial theorem but am not sure what you mean by substitute, could you show me how you would do it as I have absolutely no idea?

Answer 3

x = x, y = 1/x^2.
\[\sum^n_{k=0} \binom{n}{k} x^{n-k} y^k\]
\[\sum^9_{k=0} \binom{9}{k} x^{9-k} (\frac{1}{x^2})^k\]
\[\sum^9_{k=0} \frac{9!}{k!(9-k)!} x^{9-k} (\frac{1}{x^2})^k\]
\[\sum^9_{k=0} \frac{1(2)(3)(4)(5)(6)(7)(8)9}{k!(9-k)!} x^{9-k} (\frac{1}{x^2})^k\]
\[\sum^9_{k=0} \frac{362880}{k!(9-k)!} x^{9-k} (\frac{1}{x^2})^k\]

Answer 4

Work it out from there.

Answer 5

Do I use k = 6?

Answer 6

k=9.

Answer 7

Sorry, no. I mean n=9. You add all of the terms.

Answer 8

You add the terms for when k=0, k=1, k=2, k=3, k=4, k=5, k=6, k=7, k=8, and k=9.

Answer 9

The answers say I use 9C6 or 9C3 so I'm not really following what you're saying right now

Answer 10

It's a summation from 0 to n, where n=9. Do what you will, but n=9.

Answer 11

Yes I know that so would I do
\[\left(\begin{matrix}9 \\ 6\end{matrix}\right)\times x^{6} \times (-\frac{ 1 }{ x^{2} })^{3}\]?

Answer 12

No. It's a sum.

Answer 13

I followed your working what did I do wrong?