1. ## Binomial expansion

Could someone give me guidelines on doing this question?

2. Do you understand that $(1+x)S=S+xS?$
$S=(1-x)^9= \sum\limits_{j = 0}^9 {\binom{9}{j}\left( { - 1} \right)^j x^j }$.

3. That makes sense but I can't get the answer.

4. What can't you do? Post something of your own.

5. I've probably not added the two Sigmas properly.

6. Originally Posted by Stuck Man
I've probably not added the two Sigmas properly.
The coefficient of $x^r$ is:

$\displaystyle {9 \choose r} (-1)^r + {9 \choose r - 1} (-1)^{r-1}$.

Note:

$\displaystyle \frac{9!}{r! (9 - r)!} = \frac{9! (10 - r)}{r! (10 - r)!}$

$\displaystyle \frac{9!}{(r-1)! (10 - r)!} = \frac{9! r}{r! (10 - r)!}$

7. I have found the answer except I have 9-r instead of 10-2r.

8. I have solved it now. Thanks.

9. Is there anything wrong with this?

I don't understand why when I substitute x for 3 I get -564950498 and when I use the books answer with my calculator I get -524596891. When I substitute x for 2 the results are identical.

10. Sorry I'm talking about a different question now. I've not given the books answer for it. Anyway is there an accuracy problem with my calculator?

11. I have not included x^11 in the summation.

12. The books answer is wrong. Here is the correct answer and working.

13. I've shown that there can be an expression involving x to the power r in a sigma expression. A term containg x to the power r and not any other power of x can not be found for the rth element. The books question doesn't make sense. The book has many errors in the answers section for this group.

14. It can actually be done.

15. The book does have almost the correct answer for the first question. It should be sigma from r=0 to 10.