1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus II - Infinite Series - Geometric Series

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data


    I'm trying to solve the problem in the attachment. I was asked to evaluate the left hand side equation of the equal sign. I was unsure how to go about evaluating it so I consulted my solutions manual to look up the first step. The right hand side equation of the equal sign is what my solutions manual did for the first step. I do not see how the two are equal and what intermediate steps were left out to prove that the two are equal. I was hoping someone could explain to me what was done. I have the feeling that whatever intermediate steps were performed to go from the right hand side to the left hand side of the equal sign are very simple and is the reason why they were left hand but I can't seem to figure it out. Thanks for any help!

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Aug 15, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

  4. Aug 15, 2011 #3
    [tex]\sum\left(\frac{1}{4}\right)^k 5^{6-k} = \sum\left(\frac{1^k}{4^k}\right)\frac{5^6}{5^k} = \sum\frac{5^6}{4^k 5^k} = 5^6 \sum\frac{1}{\left(4 * 5\right)^k}= 5^6 \sum\frac{1}{20^k} = 5^6\sum\left(\frac{1}{20}\right)^k[/tex]
  5. Aug 15, 2011 #4
    Lol an algebra II thing with exponents >_>, much thanks
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Calculus Infinite Series Date
Calculus 2 - Infinite Series Nov 6, 2011
Calculus 2 - Infinite Series Nov 6, 2011
Calculus 2 - Infinite Series Nov 6, 2011
Calculus 2 infinite Series Nov 6, 2011
Calculus 2 - Infinite Series Nov 4, 2011