Moonkin
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#1
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Hi there. Stuck on this last question and finding it tricky

Find the term indepenedant of y in each of the following expansions
a) (y - 1/y)^{8}

a) (2y - 1/y^2)^{9}

All helps and hints appreciated
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Glutamic Acid
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When you have a term independent of y, the powers of y will cancel out.
Personally, I'd rewrite it as (y - y-1)^8, so for which term will these cancel out? Then find out what this value is, ie it would've been the coefficient of x for this term.

The second one is a little more tricky, but shouldn't be too difficult.
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Sir Cumference
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a) A term independent of y will need the y^a and the (-1/y)^b to cancel.
a+b=8
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Moonkin
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I still cant' solve a =(
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Square
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Right lemme see if I can help you.

You know that the binomial expansion follows:

(x+y)^n=\displaystyle \sum_{r=0}^n \binom{n}{r}(x)^{n-r}(y)^r

So for part a) You will have: (y-y^{-1})^8=\displaystyle \sum_{r=0}^8 \binom{8}{r}(y)^{8-r}(y^{-^r})

Now, from that, can you get a general expression for what each term in the binomial expansion will be?

If you do, can you think of what you would need to equate it to in order to find out the r-value for the term which is independant of y?
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Moonkin
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(Original post by Square)
Right lemme see if I can help you.

You know that the binomial expansion follows:

(x+y)^n=\displaystyle \sum_{r=0}^n \binom{n}{r}(x)^{n-r}(y)^r

So for part a) You will have: (y-y^{-1})^8=\displaystyle \sum_{r=0}^8 \binom{8}{r}(y)^{8-r}(y^{-^r})

Now, from that, can you get a general expression for what each term in the binomial expansion will be?

If you do, can you think of what you would need to equate it to in order to find out the r-value for the term which is independant of y?
Oh lawd, I'm stuck.

Heres was my thought process
(8C0) (y)^8 (y^-1)=8
(8C1) (y)^7 (y^-1)^1
...etc until (8C8)
What am I doing wrong?
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Sir Cumference
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(Original post by MilesB)
Oh lawd, I'm stuck.

Heres was my thought process
(8C0) (y)^8 (y^-1)=8
(8C1) (y)^7 (y^-1)^1
...etc until (8C8)
What am I doing wrong?
Write out the 8C4 term and see what happens.
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Square
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(Original post by notnek_01)
Write out the 8C4 term and see what happens.
That kind of ruins it :mad:
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Sir Cumference
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(Original post by Square)
That kind of ruins it :mad:
Couldn't think of any other hints to give.
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Moonkin
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(Original post by notnek_01)
Write out the 8C4 term and see what happens.
Thank you, was getting hints.
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Moonkin
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I know this is silly but in a Calculator I type for example (8C1) (y)^7 (y^-1)^1
(8C1) x (y)^7 (y^-1)^1
Is the bolded bit typed literally?
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maths-enthusiast
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could you solve the second part? it's not that different really.
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