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Math Help - Finding an. Verify my answer.

  1. #1
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    Finding an. Verify my answer.

    Hello,

    Can you please verify if my answer is correct?

    Answer:n-2/4^n
    Attached Thumbnails Attached Thumbnails Finding an. Verify my answer.-math.jpg  
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  2. #2
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    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.
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  3. #3
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    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?
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  4. #4
    Ted
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    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.
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  5. #5
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    Re: Finding an. Verify my answer.

    Quote Originally Posted by l flipboi l View Post
    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.

    (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)}.

    ....
    Where do I go from here?
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