Difference between revisions of "2017 AMC 10A Problems/Problem 1"
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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: | 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 | + | a_2 = 3 \cdot 2 + 1 = 7.\\ |
| − | a_3 = 7 | + | a_3 = 7 \cdot 2 + 1 = 15.\\ |
| − | a_4 = 15 | + | a_4 = 15 \cdot 2 + 1 = 31.\\ |
| − | a_5 = 31 | + | a_5 = 31 \cdot 2 + 1 = 63.\\ |
| − | a_6 = 63 | + | a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127} |
\end{split}</cmath>. | \end{split}</cmath>. | ||
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| + | Minor LaTeX edits by fasterthanlight | ||
== Solution 2 == | == Solution 2 == | ||
Revision as of 09:33, 28 May 2020
Problem
What is the value of
?
Solution 1
Notice this is the term
in a recursive sequence, defined recursively as
Thus:
.
Minor LaTeX edits by fasterthanlight
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
.
Solution 3
Working our way from the innermost parenthesis outwards and directly computing, we have
.
Solution 4
If you distribute this you get a sum of the powers of
. The largest power of
in the series is
, so the sum is
.
Video Solution
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.