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. :)

- anonymous

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- anonymous

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

- anonymous

for first one, did you solve
\[35k^3=21k^2\]?

- anonymous

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

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## More answers

- 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

- anonymous

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

- 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

- 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

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

27?
\

- anonymous

oops

- anonymous

\[\frac{21}{35}\]looks more like it

- anonymous

maybe even
\[\frac{3}{5}\]

- 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

- 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!

- 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

- anonymous

ick. your math teacher must hate you

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

and you want the coefficient of the term
\[x^7\]

- anonymous

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

- 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

- anonymous

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

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