## anonymous 5 years ago find the coefficient of x^6 in the expansion of(5+2x^2)^7

1. anonymous

my memory fails me on this - ill go and check

2. anonymous

It uses Mimi's Crazy theorem ;)

3. Mimi_x3

$t_{k} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k}$

4. anonymous

Mimi, that should be $$t_{k+1}$$

5. Mimi_x3

$t_{6}= \left(\begin{matrix}7 \\ 6\end{matrix}\right)5\times(2x^2)^6$

6. Mimi_x3

Nope..its only finding the coeffient

7. Mimi_x3

Not the independent term i think..

8. anonymous

You think wrong then :P

9. anonymous

its 7C3 5^4 2^3 = 35 * 625 * 8 = 175,000

10. anonymous

It's crazy right Mimi? lol

11. anonymous

where came from 7c3 5^4 2^3

12. Mimi_x3

I think that im not wrong, k+1 is used for something else..cant remember..

13. Mimi_x3

& this is not crazy, the other part is crazy ::

14. anonymous

because you have 2x^2 in the paraenthesis it will be the fourth term in the expansion so formula is 7C3 * 5^(7-3) *2^3

15. anonymous

what this teory name?

16. Mimi_x3

Binomial theorem.

17. Mimi_x3

wait..im wrong.

18. anonymous

i used the formula for the (r+1) th term of the binomial expansio (a+x)^n (r+1)th term = nC r a^(n-r) x^r

19. anonymous

Mimi, I am sure that the general term of $$(a+b)^n$$ is $$t_{k+1} = \left(\begin{matrix}n \\ k\end{matrix}\right)a^{n-k} b^{k}$$

20. anonymous

yup - thats it Fool

21. Mimi_x3

Well, in my book it says.. "Specially for the expansion of (a+b)^n , the k+1th term is...that thing..cbb typing it again