# Finding the nth term of the maclaurin's expansion of sqrt(1+x)

1. Oct 30, 2007

### rock.freak667

[SOLVED] Finding the nth term of the maclaurin's expansion of sqrt(1+x)

1. The problem statement, all variables and given/known data

Find the nth term of the Maclaurin's expansion of $\sqrt{1+x}$

2. Relevant equations

3. The attempt at a solution

so far I've expanded and gotten

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{2!4}+\frac{3x^3}{3!8}-\frac{15x^4}{4!16}+\frac{105x^5}{5!32}$$

so far I've gotten that part of the nth term should be

$$\frac{x^n}{n!2^n}$$

but what I do not know is how to deal with the + and - signs as well as the numbers in th numerators...please help me

2. Oct 30, 2007

### PowerIso

(-1)^n should cover the + and -
as for the numbers, I can't help you. I'm tired. I'll be back after I work on it.

3. Oct 30, 2007

### Dick

I think you know the pattern for the numerators. You just don't know the name for it. Look up double factorials written n!!.

4. Oct 30, 2007

### rock.freak667

But the (-1)^n doesn't work for the 2nd term or after

5. Oct 30, 2007

### Dick

You know what, I'll give you an answer. (-1)^n*(2n)!/((1-2n)*(n!)^2*4^n). I didn't work that out, I looked it up because I got annoyed with the problem. Notice the (1-2n) in the denominator changes signs between n=0 and n=1 cancelling the sign change from the (-1)^n. Getting one formula that works for all n can be more of an art than a science. I would consider it perfectly legit to write 1+x/2-x/8+... and then give a formula for the nth term after the first few. It could probably be written more simply than the one fits all version.