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    Attachment 616772I don't know how to do question 7 of this exercise. Showing that the two expressions are equal is fairly straightforward, however I have attempted the second part many times and have failed to get the correct answer. I really don't understand what the right thing to do is. Thanks for any help
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    (Original post by 11owolea)
    Attachment 616772I don't know how to do question 7 of this exercise. Showing that the two expressions are equal is fairly straightforward, however I have attempted the second part many times and have failed to get the correct answer. I really don't understand what the right thing to do is. Thanks for any help
    Let S_n=\displaystyle \sum_{r=1}^n ra^{r-1}.
    Sum over the expression on both sides over r.
    Note that the LHS is a difference of partial sums. That is to say:
    \displaystyle \sum_{r=1}^n \left(ra^r-(r-1)a^{r-1}\right)=na^n.

    The RHS is then the term that you want, plus a geometric series which you should be able to evaluate.
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    (Original post by joostan)
    Let S_n=\displaystyle \sum_{r=1}^n ra^{r-1}.
    Sum over the expression on both sides over r.
    Note that the LHS is a difference of partial sums. That is to say:
    \displaystyle \sum_{r=1}^n \left(ra^r-(r-1)a^{r-1}\right)=na^n.

    The RHS is then the term that you want, plus a geometric series which you should be able to evaluate.
    Thanks a lot, is this correct?Name:  Photo on 03-02-2017 at 18.47.jpg
Views: 23
Size:  70.8 KB
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    (Original post by 11owolea)
    Thanks a lot, is this correct?Name:  Photo on 03-02-2017 at 18.47.jpg
Views: 23
Size:  70.8 KB
    Looks good to me, also for future reference, you can check something like that on Wolfram Alpha
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    I can't see the question, but if you're asking about the sum of ra^(r-1) consider the derivative of the geometric series is also a nice way.
 
 
 
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