# rth term and general term need help with formula please

• Aug 31st 2007, 04:58 PM
lilrhino
rth term and general term need help with formula please
My problem:

Find the coefficient of the term containing $x^3$ in the expansion of $(1 - \sqrt{x})^8$

I have to find the general term and compare it with the required term.

I need to apply the rth formula in the expansion of $(a+b)^n$ to find the general term.

I figured out that n = 8 and a = 1.

What I don't understand is how the exponent for b which equals x = $\frac{1}{2}$
TIA...

See image:
• Aug 31st 2007, 07:20 PM
Jhevon
Quote:

Originally Posted by lilrhino
My problem:

Find the coefficient of the term containing $x^3$ in the expansion of $(1 - \sqrt{x})^8$

I have to find the general term and compare it with the required term.

I need to apply the rth formula in the expansion of $(a+b)^n$ to find the general term.

I figured out that n = 8 and a = 1.

What I don't understand is how the exponent for b which equals x = $\frac{1}{2}$
TIA...

See image:

what do you have to multiply 1/2 by to get 3? that is the power of the binomial expansion you have to look for, that way you can get the term where the power of x is 3

do you get it?
• Aug 31st 2007, 08:30 PM
lilrhino
Quote:

Originally Posted by Jhevon
what do you have to multiply 1/2 by to get 3? that is the power of the binomial expansion you have to look for, that way you can get the term where the power of x is 3

do you get it?

$\frac{1}{2} \div 6 = 3$

So 'r' has to be calculated before you can define the exponent for 'a'? I'm still confused as to where the $\frac{1}{2}$ comes from in the 1st place :confused:. How is that plugged into the formula?

I'm supposed to find the coefficient of the term $a^4$

I found 'r' by 10 - r + 1 = 8. 11 - 8 = 3.
• Aug 31st 2007, 08:33 PM
Jhevon
Quote:

Originally Posted by lilrhino
$\frac{1}{2} \div 6 = 3$

So 'r' has to be calculated before you can define the exponent for 'a'? I'm still confused as to where the $\frac{1}{2}$ comes from in the 1st place :confused:. How is that plugged into the formula?

I'm supposed to find the coefficient of the term $a^4$

I found 'r' by 10 - r + 1 = 8. 11 - 8 = 3.

$\sqrt {x} = x^{1/2}$

what happens if we let r = 7 in the last line of those formulas you have?

by the way, was that the formula you were given? with (r - 1) ?
• Aug 31st 2007, 08:53 PM
lilrhino
Quote:

Originally Posted by Jhevon
$\sqrt {x} = x^{1/2}$

what happens if we let r = 7 in the last line of those formulas you have?

by the way, was that the formula you were given? with (r - 1) ?

To answer the 1st question:

10 - r + 1 = 7. Then I (think) 'r' would equal 4.

To answer your 2nd question - I believe the r-1 is the exponent for 'b' and it came from the rth term formula:
• Aug 31st 2007, 09:10 PM
Jhevon
Quote:

Originally Posted by lilrhino
To answer the 1st question:

10 - r + 1 = 7. Then I (think) 'r' would equal 4.

To answer your 2nd question - I believe the r-1 is the exponent for 'b' and it came from the rth term formula:

r has to 7. why would you say 4? remember, $b$ is a term with a 1/2 power, we have to account for that. if r = 7, then r - 1 = 6 and 6*(1/2) = 3, which is the power we are after

i've always used $(a + b)^n = \sum_{r = 0}^{n} {n \choose r}a^{n - r}b^r$ ........using this formula (of course we forget about the sum, we just care about one term), we plug in r = 6 and we're home
• Aug 31st 2007, 09:15 PM
lilrhino
Quote:

Originally Posted by Jhevon
r has to 7. why would you say 4? remember, $b$ is a term with a 1/2 power, we have to account for that. if r = 7, then r - 1 = 6 and 6*(1/2) = 3, which is the power we are after

i've always used $(a + b)^n = \sum_{r = 0}^{n} {n \choose r}a^{n - r}b^r$ ........using this formula (of course we forget about the sum, we just care about one term), we plug in r = 6 and we're home

:eek: I reversed the numbers - that's how I got 4. I've never seen that formula before in any of my books. I'll study it and see if I can understand it any better than the one I'm currently using. Thanks for your detailed response :)
• Aug 31st 2007, 09:18 PM
Jhevon
Quote:

Originally Posted by lilrhino
:eek: I reversed the numbers - that's how I got 4. I've never seen that formula before in any of my books. I'll study it and see if I can understand it any better than the one I'm currently using. Thanks for your detailed response :)

it's the same formula, except you have r-1 where i have r, and our series would start at different numbers (yours would start at r = 1)