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This is probably something very easy, but wth... my mind is totally stuck right now.

I need to find the coefficient of $x^{11}$ of the expansion $(x^2 + 2\frac yx)^{10}$

Well I know that the answer is 960 from wolfram. But I cant find it on my own for some reason.

I thought I was supposed to use binomial theorem $\binom{n}ka^kb^{n-k}$ but it just doesn't work for this.

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  • $\begingroup$ What is $y$? Another variable? $\endgroup$ – user940 Dec 12 '13 at 23:34
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Multiply by $x^{10}$ to get a polynomial in $x$, namely $(x^3+2y)^{10}$, in which you are looking for the coefficient of $x^{11+10}=x^{21}$. You can treat $x^3$ as a new variable, and see that by the binomial formula this coefficient is $\binom{10}7(2y)^3=120\times2^3y^3=960y^3$, which is your answer (Wolfram notwithstanding).

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  • $\begingroup$ Thanks, but by multiplying, instead of x^11 we find x^21. The result provides the correct coefficient though. Isn't there a less complicated way to do this? I mean, is it necessary to do the multiplication part? Cheers $\endgroup$ – G. Stvns Dec 13 '13 at 15:02
  • $\begingroup$ @G.Stvns: The multiplication is just a way to free the mind from denominators. You are looking for the coefficient of $x^{11}$ in an expression that mixes positive and negative powers of $x$; the multiplication just shifts all exponents of $x$ up $10$ places so that none of them are negative any more; it is just an administarative affair. The shift means the term you are interested in now has exponent $11+10=21$ for $x$, but the coefficient, which is what interests you, has not changed. If you prefer you can apply the binomial theorem directly to $(x^2 + 2\frac yx)^{10}$. $\endgroup$ – Marc van Leeuwen Dec 13 '13 at 15:20
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Yes, it works. Now $a:=x^2$ and $b:=2y/x$. The index $k$ runs from $0$ to $n=10$, and the $k$th term has $x^{2k}/x^{n-k}\,=\,x^{2k-(n-k)}$ in it.

Can you take it from here?

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