## anonymous 4 years ago In the binomial expansion (x+k/x)^7, where k is a positive constant, the coefficients of x^3 and x are the same. Find the value of k. Using the value of k found in part (i), find the coefficient of x^7 in the expansion (1-5x^2)(x+k/x)^7. Need major help on the first part? Thanks in advance. :)

1. anonymous

$(x + \frac{k}{x})^7$ $(1 - 5x^2)(x + \frac{k}{x})^7$

2. anonymous

for first one, did you solve $35k^3=21k^2$?

3. anonymous

No, I'm completely at a lost here. Sorry. :(

4. anonymous

oh ok. well we can expand this thing all the way out if you like but maybe there is a snappier way. all the terms will look like $\dbinom{7}{j}{x^j\times(\frac{k}{x})^{n-j}}$ man that took a long time to write for some reason

5. anonymous

because in general the terms of $(a+b)^n$ look like $\dbinom{n}{j}a^jb^{n-j}$

6. anonymous

traditionally one uses "k" but you are using it for something else. now you will have 8 terms if you expand this thing, so it is not such a horrible computation, but you only want the coefficient of the $x^3$ and $x$ term

7. anonymous

we can rewrite as $\dbinom{7}{j}x^jk^{7-j}x^{j-7}$ or $\dbinom{7}{j}k^{7-j}x^{2j-7}$ i believe. when $2j-7=1$ we see $j=4$ and when $2j-7=3$ we get $j=5$ so we know what the coefficients are

8. anonymous

let me know if this makes any sense, if not i will try again.

9. anonymous

I'm sorry, but how did you get 2j - 7 = 1? D:

10. anonymous

oh because the problem said the coefficients of x^3 and x are the same. so i want to know what the coefficient of $x^1$ is. the general term looks like $x^{2j-7}$ so i set $2j-7=1$ so solve for j

11. anonymous

ok with that? that is also why i set it equal to 3

12. anonymous

OHHH. Okay, okay. I get it. Kinda slow today, sorry. And afterwards, do we just plug in the values of 'j'?

13. anonymous

i didn't actually solve this with pencil and paper so lets try it and see what we get if $j=4$ we have $\dbinom{7}{4}k^{7-4}=35k^3$ and if j = 5 we get $\dbinom{7}{5}k^{7-5}=21k^2$ so yes, i think it is right. set them equal and solve for k

14. anonymous

since presumably $k\neq 0$ the only solution is $\frac{27}{35}$

15. anonymous

27? \

16. anonymous

oops

17. anonymous

$\frac{21}{35}$looks more like it

18. anonymous

maybe even $\frac{3}{5}$

19. anonymous

btw i make no claim that this is the snappiest way to do it, there may be a much quicker method, but other than multiplying out i don't see one on the other hand we can easily check that it is right http://www.wolframalpha.com/input/?i=%28x%2Bk%2Fx%29^7

20. anonymous

I checked it on pen and paper, as well. Seems about right. :) I think I can work my way up from here. Quick way or not, it gets the job done, yeah? Thank you very, very much!

21. anonymous

yw. and i hope it is clear how to do the second part. only two ways to get $x^7$ out of your product, $(1-5x^2)(x+k/x)^7.$ $1\times x^7$ and $-5x^2\times a_5x^5$ so your real job is to find the coefficient of the term of degree 5 in the expansion you just did. i would cheat my brains out for that one

22. anonymous

ick. your math teacher must hate you

23. anonymous

No use expanding it the whole way, anyway. :)) You have no idea. Gave me 3 sets of these questions.

24. anonymous

well once you have $k=\frac{3}{5}$ and the form $\dbinom{7}{j}(\frac{3}{5})^{7-j}x^{2j-7}$ you can put $2j-7=5$ $j=6$ and compute that way.you get $a_5=\dbinom{7}{5}(\frac{3}{5}) = 21\times \frac{3}{5}$ and the multiply by -5 and then add one, (for the other x^7 term. i will try it

25. anonymous

damn i screwed up somehow. first answer is right, but this one i messed up

26. anonymous

I am... I'm sorry, but that totally confused me.

27. anonymous

oh duh, it is $\dbinom{7}{6}\times \frac{3}{5}=7\times \frac{3}{5}=\frac{21}{5}$

28. anonymous

the second part of the problem is this $(1-5x^2)(x+\frac{k}{x})^7.$ right?

29. anonymous

and you want the coefficient of the term $x^7$

30. anonymous

NONONO. I think I'd be able to do it. D: I really want to try it out by myself.

31. anonymous

so how can you get a term that has degree 7 when you multiply out? one way is you will have a term of degree seven in the expansion of the second part, times 1 and then another way ... ok i will be quiet. post if you get stuck

32. anonymous

:)) Swear I will. A million thanks to you.