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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

find the term with $x^2$

$\displaystyle\left(x^2-\frac{1}{x}\right)^{10}$

thanks!

$\displaystyle\left(x^2-\frac{1}{x}\right)^{10}$

thanks!

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

find the term with $x^2$

$\displaystyle\left(x^2-\frac{1}{x}\right)^{10}$

thanks!

$\displaystyle\left(x^2-\frac{1}{x}\right)^{10}$

thanks!

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- May 13, 2013

- 386

i did some some work here using this formula $\displaystyle a^{n-r}b^r$

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- May 13, 2013

- 386

the literal parts must be of the form $a^{n-r}b^r$ and the coefficient can be determine by using this formula $\frac{n(n-1)(n-2)...(n-r+1)}{r!}$ but i don't know how to apply this to the problem because it is a binomial in terms of x only.

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- May 13, 2013

- 386

$\displaystyle a=x^2$ $\displaystyle b=\frac{1}{x}$

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Good, so what is the general term then, based on the binomial theorem?$\displaystyle a=x^2$ $\displaystyle b=\frac{1}{x}$

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- May 13, 2013

- 386

oh no. i have no idea how to do that all i know is the term being asked is the 9th term.

because n-r= 2--->10-r=2 ---->r=8. ith term = 8+1=9th term.

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That is similar to what I have in mind. The general term is:

\(\displaystyle {10 \choose k}\left(x^2 \right)^{10-k}\left(\frac{1}{x} \right)^k\)

This can be simplified as follows:

\(\displaystyle {10 \choose k}x^{2(10-k)}\cdot x^{-k}={10 \choose k}x^{20-3k}\)

Now, we want the exponent on $x$ to be $2$, so what must the value of $k$ be?

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- May 13, 2013

- 386

k should be 6. but still i'm kind of confused of what is that "k" stand for.

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$k$ is the index of summation in the binomial theorem as I gave it above. Since $n=10$, there will be $11$ terms, for $k=0$ to $k=10$. $k=6$ represents the $7$th term.k should be 6. but still i'm kind of confused of what is that "k" stand for.

So, with $k=6$, what is this $7$th term?

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- May 13, 2013

- 386

7th term is 210x^2

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