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April 16th, 2015, 10:29 AM   #1
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Expansion to the 9th power

Hi! The problem is in the image attached. I dont know how to solve these kind of problems involving expansions beyond the 4th power.


Thanx!
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File Type: jpg Expansion to the nth power.jpg (12.8 KB, 6 views)
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April 16th, 2015, 10:44 AM   #2
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Math Focus: Calculus/ODEs
The binomial theorem tells us that:

$\displaystyle \left(ax^2-\frac{1}{ax}\right)^9=\sum_{k=0}^9\left({9 \choose k}\left(ax^2\right)^{9-k}\left(-\frac{1}{ax}\right)^k\right)=\sum_{k=0}^9\left({9 \choose k}(-1)^ka^{9-2k}x^{3(6-k)}\right)$

We see now that the term for which the exponent of $x$ is 9 has $k=3$:

$\displaystyle {9 \choose 3}(-1)^ka^{9-2\cdot3}x^{3(6-3)}$

Hence, we obtain:

$\displaystyle -84a^3=\frac{21}{2}$

Solving for $a$, we find:

$\displaystyle a^3=-\frac{1}{8}$

$\displaystyle a=-\frac{1}{2}$
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April 16th, 2015, 10:48 AM   #3
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What ever the degree, you should know that the binomial theorem says that $\displaystyle (a+ b)^n= \sum \frac{n!}{i!(n- i)!} a^ib^{n-i}$

Here, $\displaystyle a= ax^2$ and $\displaystyle b= -\frac{1}{ax}$
So the first thing you need to figure out is "what value of i makes $\displaystyle (ax^2)^i(1/ax)^{9-i}$ gives x the 9th power?
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April 16th, 2015, 12:57 PM   #4
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Quote:
Originally Posted by MarkFL View Post
The binomial theorem tells us that:

$\displaystyle \left(ax^2-\frac{1}{ax}\right)^9=\sum_{k=0}^9\left({9 \choose k}\left(ax^2\right)^{9-k}\left(-\frac{1}{ax}\right)^k\right)=\sum_{k=0}^9\left({9 \choose k}(-1)^ka^{9-2k}x^{3(6-k)}\right)$

We see now that the term for which the exponent of $x$ is 9 has $k=3$:

$\displaystyle {9 \choose 3}(-1)^ka^{9-2\cdot3}x^{3(6-3)}$

Hence, we obtain:

$\displaystyle -84a^3=\frac{21}{2}$

Solving for $a$, we find:

$\displaystyle a^3=-\frac{1}{8}$

$\displaystyle a=-\frac{1}{2}$


Thanx a lot! Also thanx for explaining the theory needed
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