• Sep 27th 2011, 06:10 PM
l flipboi l
Hello,

• Sep 27th 2011, 06:50 PM
HallsofIvy
Re: Finding an. Verify my answer.
And, in case you think I am being picky, what you give not the correct answer to either question. Perhaps if you showed how you got that we could point out errors.
• Sep 27th 2011, 07:46 PM
l flipboi l
Re: Finding an. Verify my answer.
I just needed help with finding the solution for the first part. Which is find a(subscript)n.

Here's what I got so far... I realize I did the first part wrong, here's where i'm stuck tho:

(6-n/4^n) - (6-(n+1)/4^(n+1))
(6(4^n)-n/4^n) - (6(4^(n-1)) - n + 1)/4^(n-1)

....
Where do I go from here?
• Sep 29th 2011, 12:22 PM
Ted
Re: Finding an. Verify my answer.
$\displaystyle a_n=s_n - s_{n-1}$

All what you need is your algebra.

Try to write your equations using Latex. That's will make our life easier.
• Sep 29th 2011, 01:11 PM
HallsofIvy
Re: Finding an. Verify my answer.
Quote:

Originally Posted by l flipboi l
I just needed help with finding the solution for the first part. Which is find a(subscript)n.

Here's what I got so far... I realize I did the first part wrong, here's where i'm stuck tho:

(6-n/4^n) - (6-(n+1)/4^(n+1))

No, it's the other way around. You want $\displaystyle S_{n+1}- S_n$.

Quote:

(6(4^n)-n/4^n) - (6(4^(n-1)) - n + 1)/4^(n-1)
$\displaystyle \frac{6- (n+1)}{4^{n+1}}- \frac{6- n}{4^n}$

$\displaystyle \frac{6}{4(4^n)}- \frac{n}{4(4^n)}- \frac{1}{4(4^n)}- \frac{6}{4^n}+ \frac{n}{4^n}$
$\displaystyle \frac{6}{4(4^n)}- \frac{24}{4(4^n)}- \left(\frac{n}{4(4^n)}- \frac{4n}{4(4^n}\right)- \frac{1}{4(4^n)}$.

Quote:

....
Where do I go from here?