Simplifying Summation with Logarithms

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Homework Help Overview

The discussion revolves around simplifying a summation involving logarithms, specifically the expression \(\sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})\). Participants are exploring the correctness of a derived expression and its potential for further simplification.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the summation and presents a potential answer. Some participants question the validity of the steps taken, particularly regarding the manipulation of logarithmic terms.

Discussion Status

Participants are actively engaging with the original poster's work, providing feedback on specific parts of the solution. There is a recognition of a mistake in the initial approach, and a revised understanding is emerging, although no consensus on the final form has been reached.

Contextual Notes

There is an ongoing examination of the assumptions made in the manipulation of logarithmic properties and the summation itself. Participants are also addressing the clarity of notation and the correctness of the steps involved in the simplification process.

thereddevils
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Homework Statement



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

Homework Equations





The Attempt at a Solution



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

Am i correct , or it can be further simplified ? Thanks .
 
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Hi thereddevils! :smile:

Your lg(n+1) is fine, but the two-y part isn't …

how did you manage to get a 2n outside the lg ? :wink:
 


tiny-tim said:
Hi thereddevils! :smile:

Your lg(n+1) is fine, but the two-y part isn't …

how did you manage to get a 2n outside the lg ? :wink:

Thanks , tiny tim !

oh , \sum^{n}_{r=1}\lg 2^r=\lg 2+2\lg 2+3\lg 2 +...

=\lg 2 (1+2+3+...)

=\frac{n}{2}(n+1)\lg 2

i think i see my mistake , is this correct ?
 
(have a sigma: ∑ and try using the X2 tag just above the Reply box :wink:)

Yes, that's all correct now! :smile:
 

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