Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 7"

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== Solution ==
 
== Solution ==
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Group every even term with the term following it, like so
  
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<math>(1)+(2+3)+(6+7)+(14+15)+...</math>
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Since every odd term is 1 less than a power of two, we know each group will be of the form <math>(2^{n}-2)+(2^{n}-1)</math>, or just <math>2^{n+1}-3</math>. So our sum becomes
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<math>(2^{2}-3)+(2^{3}-3)+(2^{4}-3)+...</math>
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Counting yields 500 such groups, leaving us with a geometric series minus <math>3*500=1500</math>, giving us
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<math>\frac{4*(1-2^{500})}{1-2}-1500</math>
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<math>=4*(2^{500}-1)-1500</math>
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<math>=\boxed{2^{502}-1504}</math>
  
 
== See Also ==
 
== See Also ==
 
{{UNCO Math Contest box|n=II|year=2011|num-b=6|num-a=8}}
 
{{UNCO Math Contest box|n=II|year=2011|num-b=6|num-a=8}}
 
[[Category:Intermediate Algebra Theory Problems]]
 

Latest revision as of 13:10, 17 June 2021

Problem

What is the $\underline{sum}$ of the first $999$ terms of the sequence $1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63,\cdots$ that appeared on the First Round? Recall that a term in an even numbered position is twice the previous term, while a term in an odd numbered position is one more that the previous term.

Solution

Group every even term with the term following it, like so

$(1)+(2+3)+(6+7)+(14+15)+...$

Since every odd term is 1 less than a power of two, we know each group will be of the form $(2^{n}-2)+(2^{n}-1)$, or just $2^{n+1}-3$. So our sum becomes

$(2^{2}-3)+(2^{3}-3)+(2^{4}-3)+...$

Counting yields 500 such groups, leaving us with a geometric series minus $3*500=1500$, giving us

$\frac{4*(1-2^{500})}{1-2}-1500$

$=4*(2^{500}-1)-1500$

$=\boxed{2^{502}-1504}$

See Also

2011 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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