# Finding an. Verify my answer.

• September 27th 2011, 06:10 PM
l flipboi l
Finding an. Verify my answer.
Hello,

Can you please verify if my answer is correct?

• September 27th 2011, 06:50 PM
HallsofIvy
Re: Finding an. Verify my answer.
Since the problem asks two questions, and you have only one answer, it is obvious that your answer is NOT correct.
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.
• September 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?
• September 29th 2011, 12:22 PM
Ted
Re: Finding an. Verify my answer.
$a_n=s_n - s_{n-1}$

All what you need is your algebra.

Try to write your equations using Latex. That's will make our life easier.
• September 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 $S_{n+1}- S_n$.

Quote:

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

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

Quote:

....
Where do I go from here?