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Math Help - general term

  1. #1
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    general term

    Find the general term , or nth term , u_n .

    -4 , 8 , -14 , 22 , -32

    i can see te pattern ..

    4 = 4
    8 = 4 + (2x2)
    14 = 4 + (2x2) + (2x3)
    22 = 4 + (2x2) + (2x3) + (2x4)
    32 = 4 + (2x2) + (2x3) + (2x4) + (2x5)

    But i cant do the induction ??

    i am really weak at questions like this ..
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  2. #2
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    Hi

    You can find an easy relation between |u_{n+1}| and |u_{n}| that will let you find the general expression of |u_{n}| with respect to n
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  3. #3
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    Quote Originally Posted by thereddevils View Post
    Find the general term , or nth term , u_n .

    -4 , 8 , -14 , 22 , -32

    i can see te pattern ..

    4 = 4
    8 = 4 + (2x2)
    14 = 4 + (2x2) + (2x3)
    22 = 4 + (2x2) + (2x3) + (2x4)
    32 = 4 + (2x2) + (2x3) + (2x4) + (2x5)

    But i cant do the induction ??

    i am really weak at questions like this ..
    Then u_n= 4+ (2x2)+ (2x3)+ \cdot\cdot\cdot+ (2*n) which could also be written as u_n= 4+ 2\sum_{i= 1}^n i

    Now it will help to know that \sum_{i=0}^n i= \frac{n(n-1)}{2} so that \sum_{i=1}^n i= \frac{n(n-1)}{2}- 1= \frac{n^2-n}{2}- \frac{2}{2}= \frac{n^2- n- 2}{2}
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Hi

    You can find an easy relation between |u_{n+1}| and |u_{n}| that will let you find the general expression of |u_{n}| with respect to n
    Thanks . But i am not sure how to do this .. Can u show it to me >>
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  5. #5
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    |u_1| = 4
    |u_2| = 8 = |u_1| + 4 = |u_1| + 2 \times 2
    |u_3| = 14 = |u_2| + 6 = |u_1| + 2 \times 3
    |u_4| = 22 = |u_3| + 8 = |u_3| + 2 \times 4
    ...
    |u_k| = |u_{k-1}| + 2 k


    |u_k| - |u_{k-1}| = 2 k

    Sum for k = 2 to n
    Spoiler:

    |u_n| - |u_1| = 2 \sum_{k=2}^{n} k

    |u_n| - 4 = 2 \left(\frac{n(n+1)}{2}-1\right) = n^2 + n - 2

    |u_n| = n^2 + n + 2

    And since the sign of u_n is alternatively - and +

    u_n = (-1)^{n}(n^2 + n + 2)
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  6. #6
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