Difference between revisions of "2017 AMC 10A Problems/Problem 1"

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If you distribute this you get a sum of the powers of <math>2</math>.  The largest power of <math>2</math> in the series is <math>64</math>, so the sum is <math>\boxed{\textbf{(C)}\ 127}</math>.
 
If you distribute this you get a sum of the powers of <math>2</math>.  The largest power of <math>2</math> in the series is <math>64</math>, so the sum is <math>\boxed{\textbf{(C)}\ 127}</math>.
Trump 2024
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 08:33, 5 April 2022

Problem

What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$?

$\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729$

Solution 1

Notice this is the term $a_6$ in a recursive sequence, defined recursively as $a_1 = 3, a_n = 2a_{n-1} + 1.$ Thus: \[\begin{split} a_2 = 3 \cdot 2 + 1 = 7.\\ a_3 = 7 \cdot 2 + 1 = 15.\\ a_4 = 15 \cdot 2 + 1 = 31.\\ a_5 = 31 \cdot 2 + 1 = 63.\\ a_6 = 63 \cdot 2 + 1 = \boxed{\textbf{(C)}\ 127} \end{split}\]

Minor LaTeX edits by fasterthanlight

Solution 2

Starting to compute the inner expressions, we see the results are $1, 3, 7, 15, \ldots$. This is always $1$ less than a power of $2$. The only admissible answer choice by this rule is thus $\boxed{\textbf{(C)}\ 127}$.

Solution 3

Working our way from the innermost parenthesis outwards and directly computing, we have $\boxed{\textbf{(C) } 127}$.

Solution 4

If you distribute this you get a sum of the powers of $2$. The largest power of $2$ in the series is $64$, so the sum is $\boxed{\textbf{(C)}\ 127}$.

Video Solution

https://youtu.be/str7kmcRMY8

https://youtu.be/kA6W8SwjitA

~savannahsolver

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
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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