# Binomial Theorem

• May 31st 2006, 09:00 AM
billballest
Binomial Theorem
I am having big problems with the second part of this question.

a) find a positive integer k such that

$\displaystyle (x^2)^{20-k}(\frac{1}{x})^k=x$

I have done this using the rules of powers and get k=13

b) Using the binomial theorem and the answer from a) determine the coefficient of x in the expansion of.

$\displaystyle (2x^2-\frac{1}{x})^{20}$

I don't see how the answer from part a) can help
• May 31st 2006, 09:55 AM
earboth
Quote:

Originally Posted by billballest
I am having big problems with the second part of this question.
a) find a positive integer k such that
$\displaystyle (x^2)^{20-k}(\frac{1}{x})^k=x$
I have done this using the rules of powers and get k=13

b) Using the binomial theorem and the answer from a) determine the coefficient of x in the expansion of.
$\displaystyle (2x^2-\frac{1}{x})^{20}$
I don't see how the answer from part a) can help

Hello,

If you expand this binom the first summand is $\displaystyle {20\choose 0} \cdot 2^{20} \cdot (-1)^0 \cdot x^{40}$
The power to which x is raised will decrease by 3 per summand. As you've learned in part a) you'll get a summand with a single x if you calculate the 14th summand when k = 13.

So the coefficient of x is: $\displaystyle {20\choose 13} \cdot 2^7 \cdot (-1)^{13} \cdot x = -9922560 x$

Greetings

EB
• May 31st 2006, 11:18 AM
billballest
Quote:

Originally Posted by earboth
Hello,

If you expand this binom the first summand is $\displaystyle {20\choose 0} \cdot 2^{20} \cdot (-1)^0 \cdot x^{40}$
The power to which x is raised will decrease by 3 per summand. As you've learned in part a) you'll get a summand with a single x if you calculate the 14th summand when k = 13.

So the coefficient of x is: $\displaystyle {20\choose 13} \cdot 2^7 \cdot (-1)^{13} \cdot x = -9922560 x$

Greetings

EB

Forgive my stupidity where did the $\displaystyle x^2$ go
• May 31st 2006, 01:34 PM
ThePerfectHacker
Quote:

Originally Posted by billballest
Forgive my stupidity where did the $\displaystyle x^2$ go

$\displaystyle (x^2)^{20}=x^{40}$
By one of the rule of exponents.
• May 31st 2006, 02:27 PM
billballest
$\displaystyle (2x^2-\frac{1}{x})^{20}$

Sorry I meant the $\displaystyle x^2$

In $\displaystyle 2x^2$
• May 31st 2006, 08:09 PM
earboth
Quote:

Originally Posted by billballest
$\displaystyle (2x^2-\frac{1}{x})^{20}$
Sorry I meant the $\displaystyle x^2$
In $\displaystyle 2x^2$

Hello,

as you may know a binom is the power of a sum, which is a sum of itself:

$\displaystyle (a+b)^n=\sum_{i=0}^{n}{n\choose i}\cdot a^{n-i} \cdot b^i$ = $\displaystyle {n\choose 0}\cdot a^n\cdot b^0+{n\choose 1}\cdot a^{n-1}\cdot b^1+{n\choose 2}\cdot a^{n-2}\cdot b^2$+...+$\displaystyle {n\choose n}\cdot a^{0}\cdot b^n$

So with your problem the first summands are:
$\displaystyle {20\choose 0}\cdot 2^{20} \cdot x^{2\cdot 20}\cdot (-1)^0\cdot (x^{-1})^0$ + $\displaystyle {20\choose 1}\cdot 2^{19} \cdot x^{2\cdot 19}\cdot (-1)^1\cdot (x^{-1})^1$ + $\displaystyle {20\choose 2}\cdot 2^{18} \cdot x^{2\cdot 18}\cdot (-1)^2\cdot (x^{-1})^2+...$

As you can see the exponents of the first factor decrease in steps of 2, put together with the negative exponents of the 2nd factor the complete exponent of x decreases by 3.

The highest exponent is 40. The decreasing can be described by:
Exponent = 40 - 3*i. You'll get an exponent of 1 only if i = 13 (because: 40-39 = 1)

I hope this will help a little bit further.

Greetings

EB
• Jun 1st 2006, 06:41 AM
billballest
Quote:

Originally Posted by earboth
I hope this will help a little bit further.

Greetings

EB

It did thanks very much for taking the time to explain