2011 UNCO Math Contest II Problems/Problem 7

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

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

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

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

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
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2011 UNCO Math Contest II Problems/Problem 7

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