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Math Help - summations

  1. #1
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    summations

    my teacher was in a rush at the end of the class and he literally only spent 5 minutes on summations, so i'm kinda confused here.
    Evaluate:
    10
    Σi(i^2+1)
    i=1

    Evaluate:
    10
    Σk(k- 2)(k+2)
    k=1
    oh and also if you can, i think i got it right, but i always make some dumb mistakes on these long ones

    find the derivative y = (2x^7 - x^2)(x - 1/x + 1)
    just the answer would be enough for this last one, if it doesn't match then i'll check my math again.
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  2. #2
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    Quote Originally Posted by Mr.Obson View Post
    my teacher was in a rush at the end of the class and he literally only spent 5 minutes on summations, so i'm kinda confused here.
    Evaluate:
    10
    Σi(i^2+1)
    i=1

    Evaluate:
    10
    Σk(k- 2)(k+2)
    k=1
    oh and also if you can, i think i got it right, but i always make some dumb mistakes on these long ones

    find the derivative y = (2x^7 - x^2)(x - 1/x + 1)
    just the answer would be enough for this last one, if it doesn't match then i'll check my math again.
    Love ya guys!
    Hello,

    \sum^{10}_{i=1}\left(i (i^2+1)\right) = \sum^{10}_{i=1}\left(i^3+i \right) = \sum^{10}_{i=1}\left( i^3\right) + \sum^{10}_{i=1}\left(i \right) . Use the formulae for these sums:

    \sum^{10}_{i=1}\left(i (i^2+1)\right) = \frac{10^2(10 + 1)^2}{4} + \frac12 \cdot 10(10+1) = 3080

    The second example can be done in just the same way:

    \sum^{10}_{k=1} \left(k (k+2)(k-2) \right) = \sum^{10}_{k=1}\left(k^3 - 4k\right) . Now continue.

    Unfortunately the last example isn't unambiguous:

    1. f(x)= (2x^7 - x^2) \left(x - \frac1x + 1 \right)~\longrightarrow~  f'(x)=16x^7 + 14x^6 -12x^5 -3x^2 - 2x + 1,~x \neq 0

    2. f(x)=(2x^7 - x^2)\cdot \left( \frac{x - 1}{x + 1}\right) ~\longrightarrow~ f'(x) = \frac{2x\left(7x^7+2x^6-7x^5-x^2-x+1\right)} {(x+1)^2}
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  3. #3
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    Lexington, MA (USA)
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    Hello, Mr.O!

    Evaluate: . \sum^{10}_{i=1}i(i^2+1)
    If this is the first lecture on Summations, you may not know any formulas yet.
    . . You are probably expected to find the sum directly . . .


    \sum^{10}_{i=1}i(i^2+1) \;=\;1(1^2+1)\;+\;2(2^2+1)\;+\;3(3^2+1)\;+\;4(4^2+  1) +\cdots+\;9(9^2+1)\;+\;10(10^2+1)

    . . . . . . . . = \;1\!\cdot\!2\;+\;2\!\cdot\!5\;+\;3\!\cdot\!10\;+\  ;4\!\cdot\!17\;+\;\cdots\;+\;9\!\cdot\!82\;+\;10\!  \cdot\!101

    . . . . . . . . = \;3080

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