# 1 / square root X binomial expansion

• Jun 6th 2007, 07:23 PM
geomeTRY
1 / square root X binomial expansion
I have this problem and it involves 1/square root X

I have this nagging feeling theres somthing AI have to do with it but I can't get it :(

The problem is:

Write the 1st 3 terms in each expansion in simplified form:
(x² - 1/square root X) to the power of 5
• Jun 6th 2007, 07:32 PM
Jhevon
Quote:

Originally Posted by geomeTRY
I have this problem and it involves 1/square root X

I have this nagging feeling theres somthing AI have to do with it but I can't get it :(

The problem is:

Write the 1st 3 terms in each expansion in simplified form:
(x² - 1/square root X) to the power of 5

this is Binomial expansion. there are two main ways to do it. we can use the Pascal's triangle method or the Binomial theorem method. which have you learnt?
• Jun 6th 2007, 07:40 PM
Soroban
Hello, geomeTRY!

Quote:

Write the 1st 3 terms: . $\left(x^2 -\frac{1}{\sqrt{x}}\right)^5$
I must assume you know the Binomial Theorem . . .

We have: . ${5\choose5}(x^2)^5 \;-\; {5\choose4}(x^2)^4\left(\frac{1}{\sqrt{x}}\right) \;+\; {5\choose3}(x^2)^3\left(\frac{1}{\sqrt{x}}\right)^ 2 ;+\; \cdots$

. . $= \;1\cdot x^{10} \:-\: 5\cdot x^8 \cdot \frac{1}{x^{\frac{1}{2}}} \:+\: 10 \cdot x^6\cdot\frac{1}{x}$

. . $= \;x^{10} \,-\, 5x^{\frac{15}{2}} \,+\, 10x^5$

• Jun 6th 2007, 07:41 PM
ThePerfectHacker
Quote:

Originally Posted by geomeTRY
I have this problem and it involves 1/square root X

I have this nagging feeling theres somthing AI have to do with it but I can't get it :(

The problem is:

Write the 1st 3 terms in each expansion in simplified form:
(x² - 1/square root X) to the power of 5

First do $(a+b)^5$

$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$

Now let $a=x^2 \mbox{ and } b = -\frac{1}{\sqrt{x}} = -x^{-1/2}$

We get,

$(x^2)^5 + 5(x^2)^4(-x^{-1/2}) +10 (x^2)^3 (-x^{-1/2})^2 + 10 (x^2)^2 (-x^{-1/2})^3$ $+5(x^2)(-x^{-1/2})^4 + (-x^{-1/2})^5$

Now I let you simplify that.
• Jun 6th 2007, 07:48 PM
geomeTRY
i learned it that you expand it to

tn = C(n, r) (a to the power of n-r) (b to the power of r)
• Jun 6th 2007, 07:54 PM
Jhevon
Quote:

Originally Posted by geomeTRY
i learned it that you expand it to

tn = C(n, r) (a to the power of n-r) (b to the power of r)

Yes, that's the binomial theorem:

$(a + b)^n = \sum_{r=0}^{n} {n \choose r} a^{n-r}b^r$