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Math Help - summation

  1. #1
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    summation

    \sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]

    i found the answer to be (2^n-1)\lg 2 +\lg (n+1)

    Am i correct , or it can be further simplified ? Thanks .
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  2. #2
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    Quote Originally Posted by hooke View Post
    \sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]

    i found the answer to be (2^n-1)\lg 2 +\lg (n+1)

    Am i correct , or it can be further simplified ? Thanks .

    How did you get that? Assuming lg means log, I get:

    \sum^{n}_{r=1}(\log \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\log 2^r+\log (r+1)-\log r] =\log 2\sum^n_{r=1}r+\log (n+1)=\frac{n(n+1)}{2}\,\log 2+\log (n+1)

    Tonio
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  3. #3
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    Hello hooke
    Quote Originally Posted by hooke View Post
    \sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]

    i found the answer to be (2^n-1)\lg 2 +\lg (n+1)

    Am i correct , or it can be further simplified ? Thanks .
    I agree with the final term of your answer:
    \sum^{n}_{r=1}[\lg (r+1)-\lg r]=\lg(n+1)
    But the first term is simply an AP:
    \sum^{n}_{r=1}[\lg 2^r]=\lg 2\sum^{n}_{r=1}[r]
    =\tfrac12n(n+1)\lg2
    Grandad
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