Binomial Expansion?

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Jan 15th 2009, 01:51 PM
oupib

Binomial Expansion?

Hi, I am having trouble with the following queston:

The constant in the expansions of (Equation 1) and (2) are equal, and k and p both greater than zero. Express k in terms of p.

Equation 1
( kx^2 + 6/x^2 ) ^4

Equation 2
( kx^3 + p/x^3 ) ^6

It at the end of a chapter on sequences, series and binomial expansion, I don't really understand what the question is asking me to do... I presume I should expand both equations, and that the final answer will be k=...

Why the two equations though? Are the equal to each other?
Jan 15th 2009, 02:28 PM
Soroban

Hello, oupib!

Quote:

The constant in the expansions of expression (1) and (2) are equal, and $k, p \,>\,9$

Express $k$ in terms of $p.$

Expression (1): . $\left(kx^2 + \frac{6}{x^2}\right)^4$

Expression (2): . $\left(kx^3 + \frac{p}{x^3}\right) ^6$

If we expand those binomials, we find a constant term in the middle (a term with no $x$ 's.)

In expression (1), this term is: . ${4\choose2}\left(kx^2\right)^2\left(\frac{6}{x^2}\ right)^2 \:=\:6\cdot k^2\cdot x^4\cdot\frac{36}{x^4} \;=\;216k^2$

In expression (2), this term is: . ${6\choose3}\left(kx^3\right)^3\left(\frac{p}{x^3}\ right)^3 \;=\;20\cdot k^3\cdot x^9\cdot\frac{p^3}{x^9} \;=\;20k^3p^3$

Since these terms are equal: . $20k^3p^3 \:=\:216k^2 \quad\Rightarrow\quad \frac{k^3}{k^2} \:=\:\frac{216}{20p^3}$

Therefore: . $k \;=\;\frac{54}{5p^3}$
Jan 15th 2009, 02:35 PM
oupib

Amazing! It actually makes sense now! I didn't realise that you just made them equal to each other and then simplified....

Thanks