Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 7"
(Created page with "== Problem == What is the <math>\underline{sum}</math> of the first <math>999</math> terms of the sequence <math>1, 2, 3, 6, 7, 14, 15, 30, 31, 62, 63,\cdots </math> that appear...") |
m (→Solution) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 6: | Line 6: | ||
== 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 13:10, 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 |