# Math Help - 1 / square root X binomial expansion

1. ## 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

2. 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?

3. Hello, geomeTRY!

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

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

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

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

4. 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.

5. 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)

6. 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:

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