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Math Help - Sigma notation and sums

  1. #1
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    Sigma notation and sums

    I have som question about sums never done them before, didnīt know where to post it its a analysis course so they cant be to misplaced (i hope). Am I on the right track or doing someting completely wrong? All tips appreciated.. \Rightarrow

    Write the following using Sigma notation:
    1*2*3+2*3*4+3*4*5+...+101*102*103

    \sum_{k=1}^{101}(k)(k+1)(k+2)
    2)
    1*2+2*2^2+3*2^3+...+10*2^{10}

    \sum_{k=1}^{10}k\times2^k
    3)
    then we have 1+4+9+16...+100
    \sum_{k=1}^{10}(n^k)
    and finaly
    4)
    \frac{1}{1\times 3}+\frac{1}{2\times 4}+\frac{1}{3\times 5}...

    \sum_{k=1}^{\infty} \frac{1}{k\times (k+2)}

    Actually what does this mean
    \sum_{k=1}^{n} \frac{k}{n+k} n is suppose to be the final value. The exercis said write without Sigma notation. My question is does n=k so its \frac{1}{1+1}+\frac{2}{2+2}+\frac{3}{3+3} ?
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    I have som question about sums never done them before, didnīt know where to post it its a analysis course so they cant be to misplaced (i hope). Am I on the right track or doing someting completely wrong? All tips appreciated.. \Rightarrow

    Write the following using Sigma notation:
    1*2*3+2*3*4+3*4*5+...+101*102*103

    \sum_{k=1}^{101}(k)(k+1)(k+2)
    Good.

    2)
    1*2+2*2^2+3*2^3+...+10*2^{10}

    \sum_{k=1}^{10}k\times2^k
    Good

    3)
    then we have 1+4+9+16...+100
    \sum_{k=1}^{10}(n^k)
    No. 4= 2^2, [tex]9= 32, 16= 4^2, ..., 100= 10^2
    The exponent is a constant, 2, not the index. And you don't say what "n" is!

    and finaly
    4)
    \frac{1}{1\times 3}+\frac{1}{2\times 4}+\frac{1}{3\times 5}...

    \sum_{k=1}^{\infty} \frac{1}{k\times (k+2)}
    Good! You only need to redo (2).

    Actually what does this mean
    \sum_{k=1}^{n} \frac{k}{n+k} n is suppose to be the final value. The exercis said write without Sigma notation. My question is does n=k so its \frac{1}{1+1}+\frac{2}{2+2}+\frac{3}{3+3} ?
    No. n is a fixed number and k runs from 1 to n. This is
    \frac{1}{n+1}+ \frac{2}{n+2}+ \cdot\cdot\cdot+ \frac{n-1}{n+ n-1}+ \frac{n}{n+ n}.
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