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. :)

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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. :)

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\[(x + \frac{k}{x})^7\] \[(1 - 5x^2)(x + \frac{k}{x})^7\]
for first one, did you solve \[35k^3=21k^2\]?
No, I'm completely at a lost here. Sorry. :(

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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
because in general the terms of \[(a+b)^n\] look like \[\dbinom{n}{j}a^jb^{n-j}\]
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
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
let me know if this makes any sense, if not i will try again.
I'm sorry, but how did you get 2j - 7 = 1? D:
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
ok with that? that is also why i set it equal to 3
OHHH. Okay, okay. I get it. Kinda slow today, sorry. And afterwards, do we just plug in the values of 'j'?
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
since presumably \[k\neq 0\] the only solution is \[\frac{27}{35}\]
27? \
oops
\[\frac{21}{35}\]looks more like it
maybe even \[\frac{3}{5}\]
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
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!
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
ick. your math teacher must hate you
No use expanding it the whole way, anyway. :)) You have no idea. Gave me 3 sets of these questions.
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
damn i screwed up somehow. first answer is right, but this one i messed up
I am... I'm sorry, but that totally confused me.
oh duh, it is \[\dbinom{7}{6}\times \frac{3}{5}=7\times \frac{3}{5}=\frac{21}{5}\]
the second part of the problem is this \[(1-5x^2)(x+\frac{k}{x})^7.\] right?
and you want the coefficient of the term \[x^7\]
NONONO. I think I'd be able to do it. D: I really want to try it out by myself.
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
:)) Swear I will. A million thanks to you.

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