Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 7"
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== Solution == | == Solution == | ||
| + | Group every even term with the term following it, like so | ||
| + | <math>(1)+(2+3)+(6+7)+(14+15)+...</math> | ||
| + | |||
| + | 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 | ||
| + | |||
| + | <math>(2^{2}-3)+(2^{3}-3)+(2^{4}-3)+...</math> | ||
| + | |||
| + | Counting yields 500 such groups, leaving us with a geometric series minus <math>3*500=1500</math>, giving us | ||
| + | |||
| + | <math>\frac{4*(1-2^{500})}{1-2}-1500</math> | ||
| + | |||
| + | <math>=4*(2^{500}-1)-1500</math> | ||
| + | |||
| + | <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}} | ||
Latest revision as of 12:10, 17 June 2021
Problem
What is the 
of the first 
terms of the sequence 
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
Since every odd term is 1 less than a power of two, we know each group will be of the form 
, or just 
. So our sum becomes
Counting yields 500 such groups, leaving us with a geometric series minus 
, giving us
See Also
| 2011 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
