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April 11th, 2018, 05:05 AM   #1
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Question 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.
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April 11th, 2018, 05:40 AM   #2
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Looks like the thing to try. What did you get?
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April 11th, 2018, 05:44 AM   #3
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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.
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April 11th, 2018, 06:39 AM   #4
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Quote:
Originally Posted by skipjack View Post
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.
Only if it converges at all!!
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April 11th, 2018, 06:48 AM   #5
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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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