A community for students.
Here's the question you clicked on:
 0 viewing
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 (15x^2)(x+k/x)^7.
Need major help on the first part? Thanks in advance. :)
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 (15x^2)(x+k/x)^7. Need major help on the first part? Thanks in advance. :)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(x + \frac{k}{x})^7\] \[(1  5x^2)(x + \frac{k}{x})^7\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for first one, did you solve \[35k^3=21k^2\]?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, I'm completely at a lost here. Sorry. :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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 \[\dbinom{7}{j}{x^j\times(\frac{k}{x})^{nj}}\] man that took a long time to write for some reason

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because in general the terms of \[(a+b)^n\] look like \[\dbinom{n}{j}a^jb^{nj}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0traditionally 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we can rewrite as \[\dbinom{7}{j}x^jk^{7j}x^{j7}\] or \[\dbinom{7}{j}k^{7j}x^{2j7}\] i believe. when \[2j7=1\] we see \[j=4\] and when \[2j7=3\] we get \[j=5\] so we know what the coefficients are

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let me know if this makes any sense, if not i will try again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry, but how did you get 2j  7 = 1? D:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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 \[x^{2j7}\] so i set \[2j7=1\] so solve for j

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok with that? that is also why i set it equal to 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OHHH. Okay, okay. I get it. Kinda slow today, sorry. And afterwards, do we just plug in the values of 'j'?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i didn't actually solve this with pencil and paper so lets try it and see what we get if \[j=4\] we have \[\dbinom{7}{4}k^{74}=35k^3\] and if j = 5 we get \[\dbinom{7}{5}k^{75}=21k^2\] so yes, i think it is right. set them equal and solve for k

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0since presumably \[k\neq 0\] the only solution is \[\frac{27}{35}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{21}{35}\]looks more like it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe even \[\frac{3}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0btw 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yw. and i hope it is clear how to do the second part. only two ways to get \[x^7\] out of your product, \[(15x^2)(x+k/x)^7.\] \[1\times x^7\] 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ick. your math teacher must hate you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No use expanding it the whole way, anyway. :)) You have no idea. Gave me 3 sets of these questions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well once you have \[k=\frac{3}{5}\] and the form \[\dbinom{7}{j}(\frac{3}{5})^{7j}x^{2j7}\] you can put \[2j7=5\] \[j=6\] 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0damn i screwed up somehow. first answer is right, but this one i messed up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am... I'm sorry, but that totally confused me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh duh, it is \[\dbinom{7}{6}\times \frac{3}{5}=7\times \frac{3}{5}=\frac{21}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the second part of the problem is this \[(15x^2)(x+\frac{k}{x})^7.\] right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and you want the coefficient of the term \[x^7\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0NONONO. I think I'd be able to do it. D: I really want to try it out by myself.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:)) Swear I will. A million thanks to you.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.