BTGmoderatorDC wrote:The sequence \(a_n\) is defined so that, for all \(n\) is greater than or equal to 3, \(a_n\) is the greater of \(a_{n-2} +1\) and \(a_{n-1}\). (If the two quantities are the same, then \(a_n\) is equal to either of them.) Which of the following values of \(a_1\) and \(a_2\) will produce a sequence in which no value is repeated?

A. \(a_1=-1\), \(a_2=-1.5\)
B. \(a_1=-1\), \(a_2=1\)
C. \(a_1=1\), \(a_2=-1\)
D. \(a_1=1\), \(a_2=1.5\)
E. \(a_1=1.5\), \(a_2=1\)

Source: Manhattan Prep

\[{a_n} = \max \left( {{a_{n - 2}} + 1\,\,;\,\,{a_{n - 1}}} \right)\,\,\,;\,\,\,n \geqslant 3\]
(When both coincide, max - by definition - equals any one of them, exactly as the question stem imposes!)

In English: (*) Each term (from the third on) is equal to the greatest between the previous AND one more than the "previous of the previous".

What is in blue is ESSENCIAL, because it helps you to do VERY quickly everything below!

\[?\,\,\,:\,\,\,{\text{no}}\,\,{\text{repetitions}}\]

In "Which of the following..." problems, we usually must test the alternatives.
In this case, we see no reason for doing that in an "unnatural order", therefore we go "natural", i.e., from (A) to (E).
\[\left( A \right)\,\,\,\left. \begin{gathered}
{a_1} = - 1 \hfill \\
{a_2} = - 1.5 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_3} = \max \left( { - 1.5\,;\,0} \right) = 0\]
\[\left. \begin{gathered}
{a_2} = - 1.5 \hfill \\
{a_3} = \boxed0 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_4} = \max \left( {0\,;\, - 0.5} \right) = \boxed0\]

\[\left( B \right)\,\,\,\left. \begin{gathered}
{a_1} = - 1 \hfill \\
{a_2} = \boxed1 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_3} = \max \left( {1\,;\,0} \right) = \boxed1\]

\[\left( C \right)\,\,\,\left. \begin{gathered}
{a_1} = 1 \hfill \\
{a_2} = - 1 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_3} = \max \left( { - 1\,;\,2} \right) = 2\]
\[\left. \begin{gathered}
{a_2} = - 1 \hfill \\
{a_3} = \boxed2 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_4} = \max \left( {2\,;\,0} \right) = \boxed2\]

\[\left( D \right)\,\,\,\left. \begin{gathered}
{a_1} = 1 \hfill \\
{a_2} = 1.5 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_3} = \max \left( {1.5\,;\,2} \right) = 2\]
\[\left. \begin{gathered}
{a_2} = 1.5 \hfill \\
{a_3} = 2 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_4} = \max \left( {2\,;\,2.5} \right) = 2.5\]
\[\left. \begin{gathered}
{a_3} = 2 \hfill \\
{a_4} = 2.5 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_5} = \max \left( {2.5\,;\,3} \right) = 3\]
\[\left. \begin{gathered}
{a_4} = 2.5 \hfill \\
{a_5} = 3 \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_6} = \max \left( {3\,;\,3.5} \right) = 3.5\]

By "vulgar induction" it is not hard to "understand" (or "conclude") that we will have:
\[{a_7} = 4\,\,\,\,;\,\,\,\,{a_8} = 4.5\,\,\,\,;\,\,\,\,{a_9} = 5\,\,\,\,;\,\,\,\, \ldots \]
(No repetitions, therefore (D) is the answer we were looking for!)

Important: if you feel insecure, test alternative choice (E) to confirm there is repetition.
Conclusion: (D) will be the right answer by EXCLUSION.

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

BTGmoderatorDC wrote:The sequence \(a_n\) is defined so that, for all \(n\) is greater than or equal to 3, \(a_n\) is the greater of \(a_{n-2} +1\) and \(a_{n-1}\). (If the two quantities are the same, then \(a_n\) is equal to either of them.) Which of the following values of \(a_1\) and \(a_2\) will produce a sequence in which no value is repeated?

A. \(a_1=-1\), \(a_2=-1.5\)
B. \(a_1=-1\), \(a_2=1\)
C. \(a_1=1\), \(a_2=-1\)
D. \(a_1=1\), \(a_2=1.5\)
E. \(a_1=1.5\), \(a_2=1\)

OA D

Source: Manhattan Prep

Testing values seems laborious and time consuming under the circumstances.

So if you assume that an ever increasing sequence won't yield duplicates, then a2>a1, so that would eliminate A, C and E, leaving only B and D

Now under the vague idea that a sequence of negative and positive numbers are more likely to yield duplicates, that would eliminate B, leaving D as a plausible answer.

regor60 wrote:Testing values seems laborious and time consuming under the circumstances.

So if you assume that an ever increasing sequence won't yield duplicates, then a2>a1, so that would eliminate A, C and E, leaving only B and D

Now under the vague idea that a sequence of negative and positive numbers are more likely to yield duplicates, that would eliminate B, leaving D as a plausible answer.

Well observed and put!

POST-MORTEM: a close attention to the "law of formation of the sequence":
\[{a_n} = \max \left( {{a_{n - 2}} + 1\,\,;\,\,{a_{n - 1}}} \right)\,\,\,;\,\,\,n \geqslant 3\]
and keeping in mind our FOCUS ("avoid repetitions") it is EASY to see that we want, starting with the third element of the sequence, that:
\[{a_{n - 2}} + 1\,\, > {a_{n - 1}}\,\,\,\,\,\,\left( {{\text{to}}\,\,{\text{avoid}}\,\,{a_n} = {a_{n - 1}}} \right)\]
In English: each term "a_(n-1)" must be less than its previous plus one ("a_(n-2)+1") !

Important: this "English translation" usually helps a LOT, not only saving time but also avoiding "silly mistakes" in substitutions.

Regards,
fskilnik.