Author Topic: Binomial theorem (Read 1940 times) Tweet Share

Aqualim · « **Reply #15 on:** January 05, 2010, 07:26:26 pm »

Just to shed some light on the question you were asking about whether there was an eaiser way of solving this kind of equation, well there is!

It's probably already been explained to you in some form, but I'll share it with you anyway.

So basically we know that when finding coefficents and the 'x' value is on the left side of the question e.g. $(x-5)^5$ we count the terms in a descending order, meaning from $x^5 \to x^0$ and when the 'x' value is on the right side we count the terms in an ascending order $x^0 \to x^5$ .

Using this idea, we can do questions having two 'x' values. E.g.

$\left (x^2 \right )^6 \left ( -\frac{3}{x} \right )^0 = x^{12}$

$\left (x^2 \right )^5 \left ( -\frac{3}{x} \right )^1 = x^9$

$\left (x^2 \right )^4 \left ( -\frac{3}{x} \right )^2 = x^6$

$\left (x^2 \right )^3 \left ( -\frac{3}{x} \right )^3 = x^3$

$\left (x^2 \right )^2 \left ( -\frac{3}{x} \right )^4 = x^0$

$\left (x^2 \right )^1 \left ( -\frac{3}{x} \right )^5 = \frac{1}{x^3}$

$\left (x^2 \right )^0 \left ( -\frac{3}{x} \right )^6 = \frac{1}{x^6}$

NOTE: We are only interested in the 'x' values $\therefore$ when subracting the right side from the left, we don't include them, hence ending up with an 'x' value.

I hope that made some sense

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Author Topic: Binomial theorem (Read 1940 times) Tweet Share

Aqualim

Re: Binomial theorem

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