Serious Series

Algebra Level 4

Let the sum of all terms of the infinite series 2 , 3 4 , 5 16 , 9 64 , 17 256 , 2,\frac{3}{4},\frac{5}{16},\frac{9}{64},\frac{17}{256},\cdots , be x x .

Let x x have a rational representation as x = a b x=\frac{a}{b} , where a a and b b are co-prime.

What is the value of a + b a+b .


The answer is 13.

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3 solutions

The series can be written as

1 + 1 , 2 + 1 4 , 2 2 + 1 4 2 , 2 3 + 1 4 3 , 1+1,\frac{2+1}{4},\frac{2^2+1}{4^2},\frac{2^3+1}{4^3},\cdots

which can be rewritten as

1 + 1 , 1 2 + 1 4 , ( 1 2 ) 2 + ( 1 4 ) 2 , ( 1 2 ) 3 + ( 1 4 ) 3 , 1+1,\frac{1}{2}+\frac{1}{4},\left(\frac{1}{2}\right)^2+\left(\frac{1}{4}\right)^2,\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^3,\cdots

The series is the sum of terms of two convergent GPs both starting at 1 and with ratios 1 2 \frac{1}{2} and 1 4 \frac{1}{4} .

Thus, the required sum would be

S = 1 1 1 2 + 1 1 1 4 = 2 + 4 3 = 10 3 S = \frac{1}{1-\frac{1}{2}}+\frac{1}{1-\frac{1}{4}}=2+\frac{4}{3}=\frac{10}{3}

Thus, a = 10 , b = 3 , a + b = 13 a=10,b=3,a+b=\boxed{13}

Sridhar Sri
Feb 23, 2016

That is a beautiful approach. I just loved it.

Soumava Pal - 5 years, 3 months ago

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Thank you @Soumava Pal

Sridhar Sri - 5 years, 3 months ago

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You are welcome. I did it by @Janardhanan Sivaramakrishnan 's method only, but yours seemed a lot elegant.

Soumava Pal - 5 years, 3 months ago

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@Soumava Pal Also we can do it by dividing by 4.. and subtracting ..But both come same after 2 or 3 steps.

Sridhar Sri - 5 years, 3 months ago

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@Sridhar Sri Yes. @Victor Victal 's solution.

Soumava Pal - 5 years, 3 months ago
Vitor Victal
Jan 19, 2015

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