Question

Find the coefficient of x^2 in the expansion of:

(x-1/x)^9*(x+1/x)^5

Answer 1

is that (x-1)/x and (x+1)/x or x-(1/x) and x+(1/x)? @Mathematique

Answer 2

@Mathematique hello are you there?

Answer 3

Note that:\[\left(x-\frac{1}{x}\right)^9=\sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)x^{9-k}\left(-\frac{1}{x}\right)^k\]\[=\sum_{k=0}^9 (-1)^k \left(\begin{matrix}9 \\ k\end{matrix}\right)x^{9-2k}\]Similarly:\[\left(x+\frac{1}{x}\right)^5=\sum_{l=0}^5\left(\begin{matrix}5 \\ l\end{matrix}\right)x^{5-2l}.\]Hence:\[\left(x-\frac{1}{x}\right)^9\cdot \left(x+\frac{1}{x}\right)^5\]\[=\left(\sum_{k=0}^9 (-1)^k \left(\begin{matrix}9 \\ k\end{matrix}\right)x^{9-2k}\right)\left(\sum_{l=0}^5\left(\begin{matrix}5 \\ l\end{matrix}\right)x^{5-2l}\right).\]As we let k range between 0 and 9, and l range between 0 and 5, we need to see when:\[(9-2k)+(5-2l)=2\Longrightarrow14-2(k+m)=2 \]\[\Longrightarrow -2(k+m)=-12\Longrightarrow k+m=6\Longrightarrow m=6-k\]This happens when:\[k=1, m=5\]\[k=2, m=4\]\[k=3, m=3\]\[k=4, m=2\]\[k=5, m=1\]\[k=6, m=0.\]Thus the coefficient in front of the x^2 will be:\[\sum_{k=1}^6 (-1)^k \left(\begin{matrix}9 \\ k\end{matrix}\right)\left(\begin{matrix}5 \\ 6-k\end{matrix}\right).\]

Answer 4

just noticed, for some reason I switched from l to m in the middle of that <.< my mistake >.< (was copying from a piece of paper)

Answer 5

I dont think the problem is:\[\left(\frac{x-1}{x}\right)^9\]but rather:\[\left(x-\frac{1}{x}\right)^9.\] Until the OP comes back to say which is correct we will never really know what the problem is.

Answer 6

Regardless, considering what makes sense to me, I will assume that it's x-(1/x) and x+(1/x). Then we can expand each binomial through the binomial theorem. Hence the binomial expansion for each of the expressions is like this:\[\bf \left( x-\frac{1}{x} \right)^9=\sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)(x)^{9-k}\left( \frac{ 1 }{ x } \right)^k(-1)^k= \sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)(x)^{9-2k}(-1)^k = \]\[\bf x^9-9x^7+36x^5-84x^3+126x-126x^{-1}+84x^{-3}-36x^{-5}+9x^{-7}-x^{-9}\]\[\bf \left( x+\frac{1}{x} \right)^9=\sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)(x)^{9-k}\left( \frac{ 1 }{ x } \right)^k= \sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)(x)^{9-2k} = \]\[\bf x^9+9x^7+36x^5+84x^3+126x+126x^{-1}+84x^{-3}+36x^{-5}+9x^{-7}+x^{-9} \]Let the latter binomial expansion be denoted by \(\bf A(x)\). It is clear to see that when the two expansions get multiplied together through the distributive property, the product will be given like the following:\[\bf x^9A(x)-9x^7A(x)+36x^5A(x)-84x^3A(x)+126xA(x)-126x^{-1}A(x)\]\[\bf +84x^{-3}A(x)-36x^{-5}A(x)+9x^{-7}A(x)-x^{-9}A(x)\]In each part of the product containing A(x), the only way an \(\bf x^2\) is produced is when the exponent of one of the x's in \(\bf A(x)\) and the exponent of the term being multiplied add up to 2. For example: \(\bf x^9\) produces a \(\bf 9x^2\) when multiplied b y the \(\bf 9x^{-7}\) in \(\bf A(x)\). Continuing this in a similar manner, the coefficient of \(\bf x^2\) will be given as : \[\bf (9)x^2+(-9+36)x^2+(36+84)x^2+...+(9)x^2=510x^2\]Hence the coefficient of \(\bf x^2\) in the expansion is \(\bf 510\).

Answer 7

@joemath314159 You get same answer as me?

Answer 8

@Loser66 hbu?

Answer 9

You calculated \[\left(x+\frac{1}{x}\right)^9\] but the problem has \[\left(x+\frac{1}{x}\right)^5\] >.<

Answer 10

OMFGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

Answer 11

@joemath314159 i hate u. lol.

Answer 12

Regardless. Had the problem been what I thought it was =.=, the answer I obtained should be correct.

Answer 13

I give you a medal because the method is correct :)

Answer 14

Should I redo it with the power of 5? Shouldn't take too long :D And how about we both calculate answers and let's see if they're the same?

Answer 15

Sure, I got 45 I think. Let me double check.

Answer 16

\[(-1)^1\left(\begin{matrix}9 \\ 1\end{matrix}\right)\left(\begin{matrix}5 \\ 5\end{matrix}\right)+(-1)^2\left(\begin{matrix}9 \\ 2\end{matrix}\right)\left(\begin{matrix}5 \\ 4\end{matrix}\right)+(-1)^3\left(\begin{matrix}9 \\ 3\end{matrix}\right)\left(\begin{matrix}5 \\ 3\end{matrix}\right)\]\[+(-1)^4\left(\begin{matrix}9 \\ 4\end{matrix}\right)\left(\begin{matrix}5 \\ 2\end{matrix}\right)+(-1)^5\left(\begin{matrix}9 \\ 5\end{matrix}\right)\left(\begin{matrix}5 \\ 1\end{matrix}\right)+(-1)^6\left(\begin{matrix}9 \\ 6\end{matrix}\right)\left(\begin{matrix}5 \\ 0\end{matrix}\right)\]\[=-9+180-840+1260-630+84\]\[=45\]

Answer 17

verified by wolframalpha: http://www.wolframalpha.com/input/?i=coefficient+of+x%5E2+in+%28x-1%2Fx%29%5E9%28x%2B1%2Fx%29%5E5

Answer 18

\[\bf \left( x-\frac{1}{x} \right)^9=\sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)(x)^{9-k}\left( \frac{ 1 }{ x } \right)^k(-1)^k=\sum_{k=0}^{9}\left(\begin{matrix}9 \\ k\end{matrix}\right)(x)^{9-2k}(-1)^k = \]\[\bf x^9-9x^7+36x^5-84x^3+126x-126x^{-1}+84x^{-3}-36x^{-5}+9x^{-7}-x^{-9}\]\[\bf \left( x+\frac{1}{x} \right)^5=\sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)(x)^{5-k}\left( \frac{ 1 }{ x } \right)^k= \sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)(x)^{5-2k} = \]\[\bf x^5+5x^3+10x+10x^{-1}+5x^{-3}+x^{-5}\]
Let the latter binomial expansion be denoted by \(\bf A(x)\). It is clear to see that when the two expansions get multiplied together through the distributive property, the product will be given like the following:\[\bf x^9A(x)-9x^7A(x)+36x^5A(x)-84x^3A(x)+126xA(x)-126x^{-1}A(x)\]\[\bf +84x^{-3}A(x)-36x^{-5}A(x)+9x^{-7}A(x)-x^{-9}A(x)\]In each part of the product containing A(x), the only way an \(\bf x^2\) is produced is when the exponent of one of the x's in \(\bf A(x)\) and the exponent of the term being multiplied add up to 2. For example, \(\bf -9x^7\) produces a \(\bf -9x^2\) when multiplied by \(\bf x^{-5}\) in \(\bf A(x)\). Continuing this in a similar manner, the coefficient of \(\bf x^2\) will be given as:\[(-9)x^2+(36 \times 5)x^2+(-84 \times 10)x^2+(126 \times 10)x^2+(-126 \times 5)x^2+(84)x^2=\bf 45x^2 \]Hence the coefficient of \(\bf x^2\) in the expansion is \(\bf 45\).

@joemath314159 Same as you =]
@Loser66
@Mathematique

Answer 19

Haha sorry for the late reply! Yes I did manage to solve it! Thanks to both of you for your feedback, and yes it was 1+(1/x) :)