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

(Solution to problem 1)
(Solution)
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
Notice this is <math>a_6</math> arithmetic sequence <math>a_1 = 3, a_n = 2a_(n-1) + 1.</math>\\
+
Notice this is the term <math>a_6</math> in an arithmetic sequence, such that:
<math>a_2 = 3*2 + 1 = 7.</math>\\
+
<cmath>\begin{split}
<math>a_3 = 7 *2 + 1 = 15.</math>\\
+
a_1 = 3.\\
<math>a_4 = 15*2 + 1 = 31.</math>\\
+
a_n = 2a_(n-1) + 1.\\
<math>a_5 = 31*2 + 1 = 63.</math>\\
+
It follows that:\\
<math>a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127.}</math>
+
a_2 = 3*2 + 1 = 7.\\
 +
a_3 = 7 *2 + 1 = 15.\\
 +
a_4 = 15*2 + 1 = 31.\\
 +
a_5 = 31*2 + 1 = 63.\\
 +
a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127.}
 +
\end{split}</cmath>

Revision as of 14:20, 8 February 2017

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

Notice this is the term $a_6$ in an arithmetic sequence, such that: \[\begin{split} a_1 = 3.\\ a_n = 2a_(n-1) + 1.\\ It follows that:\\ a_2 = 3*2 + 1 = 7.\\ a_3 = 7 *2 + 1 = 15.\\ a_4 = 15*2 + 1 = 31.\\ a_5 = 31*2 + 1 = 63.\\ a_6 = 63*2 + 1 = \boxed{\textbf{(C)}\ 127.} \end{split}\]