April 11th, 2018, 05:05 AM  #1 
Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0  Infinite Sums
T = 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 The numerator of each subsequent term is the sum of the numerators of the previous two terms. The denominator of each subsequent term is twice the denominator of the previous term. 1) By considering the first six terms of 1/2 T and 1/4 T, find the values of a, b, c, d, e and f where 3/4 T = a/4+ b/8 + c/16 + d/32 + e/64 + f/128. 2) Hence find the value of T. For 1) I tried doing 1/2 T and 1/4 T and adding them together, but I am not sure whether that is the correct method. Please help! Thanks. Last edited by skipjack; April 11th, 2018 at 05:46 AM. 
April 11th, 2018, 05:40 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2455 Math Focus: Mainly analysis and algebra 
Looks like the thing to try. What did you get?

April 11th, 2018, 05:44 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,174 Thanks: 1644 
T $\ \ \ \ $ = 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + . . . T/2 $\ $ = 1/4 + 1/8 + 2/16 + 3/32 + 5/64 + 8/128 + . . . T/4 $\ $ = $\ \ \ \ \ $ 1/8 + 1/16 + 2/32 + 3/64 + 5/128 + 8/256 + . . . 3T/4 = 1/4 + 2/8 + 3/16 + 5/32 + 8/64 + 13/128 + . . . = T  1/2 Hence T = 2. 
April 11th, 2018, 06:39 AM  #4  
Senior Member Joined: Oct 2009 Posts: 428 Thanks: 144  Quote:
 
April 11th, 2018, 06:48 AM  #5 
Senior Member Joined: Oct 2009 Posts: 428 Thanks: 144 
A method that also shows convergence is to represent the numerators (the Fibonnaci numbers) by their formula: $$\sum \frac{F_0}{2^n} = \sum \frac{\varphi^n  (\varphi)^{n}}{2^n \sqrt{5}} = \frac{1}{\sqrt{5}}\sum \left(\frac{\varphi}{2}\right)^n  \frac{1}{\sqrt{5}} \sum\left(\frac{1}{2\varphi}\right)^n$$ Using geometric series, we get $$\frac{1}{\sqrt{5}}\frac{1}{1 (\varphi/2)}  \frac{1}{\sqrt{5}}\frac{1}{1+(1/(2\varphi)} $$ Some tedious algebra shows that this indeed evaluates to 2. 

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