Hello,

Can you please verify if my answer is correct?

Answer:n-2/4^n

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- Sep 27th 2011, 06:10 PMl flipboi lFinding an. Verify my answer.
Hello,

Can you please verify if my answer is correct?

Answer:n-2/4^n - Sep 27th 2011, 06:50 PMHallsofIvyRe: 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. - Sep 27th 2011, 07:46 PMl flipboi lRe: 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 PMTedRe: 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 PMHallsofIvyRe: Finding an. Verify my answer.
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)

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

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

Quote:

....

Where do I go from here?