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anonymous

  • 5 years ago

Write the nth term of the arithmetic sequence as a function of n. a[1]=8, a[k+1]=a[k]+7

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  1. anonymous
    • 5 years ago
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    The formula for the nth term of an arithmetic sequence is a[n] = a[1]+(n-1)d. You know that d= a[k+1]-a[k] = 7, and a[1]=8, so the nth term is a[n] = 8 + 7(n-1)

  2. anonymous
    • 5 years ago
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    ok so I was close but the problem is that none of the answers I can choose from match with that...let me show you my choices...

  3. anonymous
    • 5 years ago
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    a. a[n]=15+7n b. a[n]=1+7n c. a[n]=8+7n d. a[n]=1+7(n-1) e. a[n]= -1+8n

  4. anonymous
    • 5 years ago
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    If you expand the solution i sent, you'll find a[n] = 8 +7n - 7 = 1 + 7n...answer b.

  5. anonymous
    • 5 years ago
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    oh oh ok sorry about that after I posted it I figured that out thank you and I am your fan now:)

  6. anonymous
    • 5 years ago
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    No worries.

  7. anonymous
    • 5 years ago
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    do you understand how to use the sigma notation?

  8. anonymous
    • 5 years ago
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    Yes.

  9. anonymous
    • 5 years ago
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    Use sigma notation to write the sum. (1/3.2)+(1/4.3)+...+(1/7.6)

  10. anonymous
    • 5 years ago
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    a. \[\sum_{n=1}^{5}\] (1/(n+1)(n+2))

  11. anonymous
    • 5 years ago
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    It will be sum from n=3 to 7 of (10/11n-1).

  12. anonymous
    • 5 years ago
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    b. \[\sum_{n=1}^{5}(1/(n(n+1)))\]

  13. anonymous
    • 5 years ago
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    ok thank you!

  14. anonymous
    • 5 years ago
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    Do you need an explanation?

  15. anonymous
    • 5 years ago
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    Yes that would be helpful... I mean I kinda get it but it wouldn't hurt anything

  16. anonymous
    • 5 years ago
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    The denominator is in the form of some number n + (n-1)/10 (e.g. 3.2 = 3 + (3-1)/10). You then just take the reciprocal of n+(n-1)/10 and use sigma notation for n = 3 to 7.

  17. anonymous
    • 5 years ago
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    I just cleaned up the fraction before using sigma.

  18. anonymous
    • 5 years ago
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    oh ok makes it sound alot easier thanks!

  19. anonymous
    • 5 years ago
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    \[\sum_{n=3}^{7}1/[n+(n-1)/10]\]

  20. anonymous
    • 5 years ago
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    Find the sum of the infinite series. \[\sum_{i=1}^{\infty}4(-1/4)^i\] a.undefined b.-4/3 c.4 d.8/5 e.-4/5

  21. anonymous
    • 5 years ago
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    I'm pretty sure the answer's -4/5, but I want to check something.

  22. anonymous
    • 5 years ago
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    okay...

  23. anonymous
    • 5 years ago
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    Yes...I don't know what level of maths you're doing, but first you need to ensure that the series is convergent. Since it's absolutely convergent, it will be convergent. It also means you can split the sum up into the positive and negative components and use the formula for the limiting value in a geometric series to determine each of the sums. You then end up with -4/5. Do you know how to do this?

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