View source for 2011 UNCO Math Contest II Problems/Problem 7

== 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 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

<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 ==
{{UNCO Math Contest box|n=II|year=2011|num-b=6|num-a=8}}