Thefirst four termsin the expansion of $\displaystyle (1+px)^{n}$, where $\displaystyle n>0$ are $\displaystyle 1+qx+66k^{2}x^{2}+5940x^{3}$. Calculate the value ofn, ofpand ofq.

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- Nov 14th 2007, 06:17 PMacc100jtBinomial Theorem
The

**first four terms**in the expansion of $\displaystyle (1+px)^{n}$, where $\displaystyle n>0$ are $\displaystyle 1+qx+66k^{2}x^{2}+5940x^{3}$. Calculate the value of*n*, of*p*and of*q*.

Thanks! - Nov 14th 2007, 06:39 PMThePerfectHacker
- Nov 14th 2007, 06:52 PMTruthbetold
Never seen that kind of problem before.

Does it go on? or is that the end at 5940? I'll assume it does end at the highest exponent 3.

Where does k come from? I'll assume it doesn't matter or it's p, which can be solved after we know what p is.

For this,

a= 1

b= px

because of the readiness to plug it into pascal's triangle as a and b.

Pascal's triangle. $\displaystyle [(a+b)^3= a^3 + 3a^2*b + 3a*b^2 + b^3$

The problem ends with the highest power being 3 (on $\displaystyle x^3$ ) which fits the the above part of Pascal's triangle.

So we now have n.

So what to the 3rd power equals? 5940

$\displaystyle p^n= 5940$

So now we have p.

$\displaystyle 3a^2*b$.... is the place where qx exists.

So q= 3(Px)

(because 1 which is a to the second power (1^2) is just one and it doesn't change the equation)

Because p= a in its location in pascal's triangle

and solve.

phew!

Ask if I talked too complicated or missed a mental/verbal/mathematical/grammatical/social/economical :) step. - Nov 14th 2007, 06:54 PMacc100jt
Take note that $\displaystyle n$ is not 3. The question says "first four terms".

*n*can be any number.

Anyway, the answer is $\displaystyle n=12, p=3, q=36$

Can anyone show me how to get the answer? - Nov 15th 2007, 03:04 AMPlato