Difference between revisions of "2017 AMC 10A Problems/Problem 1"
Dacalculus (talk | contribs) (→Solution: The sequence is not an arithmetic sequence) |
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− | ==Solution== | + | ==Solution 1== |
− | Notice this is the term <math>a_6</math> in a recursive sequence, defined recursively as <math>a_1 = 3, a_n = 2a_ | + | Notice this is the term <math>a_6</math> in a recursive sequence, defined recursively as <math>a_1 = 3, a_n = 2a_{n-1} + 1.</math> Thus: |
<cmath>\begin{split} | <cmath>\begin{split} | ||
a_2 = 3*2 + 1 = 7.\\ | a_2 = 3*2 + 1 = 7.\\ | ||
Line 14: | Line 14: | ||
a_4 = 15*2 + 1 = 31.\\ | a_4 = 15*2 + 1 = 31.\\ | ||
a_5 = 31*2 + 1 = 63.\\ | a_5 = 31*2 + 1 = 63.\\ | ||
− | a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127 | + | a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127} |
\end{split}</cmath> | \end{split}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Starting to compute the inner expressions, we see the results are <math>1, 3, 7, 15, \ldots</math>. This is always <math>1</math> less than a power of <math>2</math>. The only admissible answer choice by this rule is thus \boxed{\textbf{(C)}\ 127}. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2017|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:52, 8 February 2017
Contents
Problem
What is the value of ?
Solution 1
Notice this is the term in a recursive sequence, defined recursively as Thus:
Solution 2
Starting to compute the inner expressions, we see the results are . This is always less than a power of . The only admissible answer choice by this rule is thus \boxed{\textbf{(C)}\ 127}.
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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