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Algebra Level 2

If the value of ∑ n = 1 16 1 2 n \displaystyle \sum_{n=1}^{16} \dfrac1{2^n} can be expressed as a b \dfrac ab , where a a and b b are coprime positive integers , find a + b a+b .


The answer is 131071.

Lee Care Gene by Lee Care Gene , 20, Malaysia

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1 solution

Lee Care Gene
Aug 31, 2016

∑ 16 1 1 2 n = ( 1 2 1 + 1 2 2 + 1 2 3 + ⋯ + 1 2 14 + 1 2 15 + 1 2 16 + 1 2 16 ) − 1 2 16 = 1 − 1 65536 = 65535 65536 \displaystyle\sum _{ 16 }^{ 1 }{ \frac { 1 }{ { 2 }^{ n } } } \\ =\left( \frac { 1 }{ { 2 }^{ 1 } } +\frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 3 } } +\cdots +\frac { 1 }{ { 2 }^{ 14 } } +\frac { 1 }{ { 2 }^{ 15 } } +\frac { 1 }{ { 2 }^{ 16 } } +\frac { 1 }{ { 2 }^{ 16 } } \right) -\frac { 1 }{ { 2 }^{ 16 } } \\ =1-\frac { 1 }{ 65536 } \\ =\frac { 65535 }{ 65536 }

a + b = 65535 + 65536 = 131071 a+b =65535+65536 =131071

4 Helpful 0 Interesting 0 Brilliant 0 Confused

It will be much better if you ask b-a.

Sabhrant Sachan - 4 years, 9 months ago

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That is true. I just realized that :).

Lee Care Gene - 4 years, 9 months ago

Why did you subtract 2^(-16)?

Anandmay Patel - 4 years, 9 months ago

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I had added another 2^(-16) in the bracket to make it sum up as 1 so i need to subtract it again.

Lee Care Gene - 4 years, 9 months ago

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oh yes...How i missed it?!?!?!Thanks.

Anandmay Patel - 4 years, 9 months ago

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