AMC 10 · 2008 · #11
Grade 8 algebraSuppose that (un) is a sequence of real numbers satifying un+2=2un+1+un,
and that u3=9 and u6=128. What is u5?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A sequence follows the rule $u_{n+2}=2u_{n+1}+u_n$: each term is twice the term before it plus the term before that. Given $u_3=9$ and $u_6=128$, find $u_5$.
Givens: The rule $u_{n+2}=2u_{n+1}+u_n$ holds for every $n$.; $u_3=9$.; $u_6=128$.
Unknowns: The value of $u_5$.
Understand
Restated: A sequence follows the rule $u_{n+2}=2u_{n+1}+u_n$: each term is twice the term before it plus the term before that. Given $u_3=9$ and $u_6=128$, find $u_5$.
Givens: The rule $u_{n+2}=2u_{n+1}+u_n$ holds for every $n$.; $u_3=9$.; $u_6=128$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities
The gap between the known term $u_3$ and the known term $u_6$ is filled by two hidden terms, $u_4$ and $u_5$. Tool #4 (Introduce a Variable) names those two unknowns so we can work with them. Tool #13 (Convert to Algebra) then turns the recurrence into equations: applying the rule once gives a relation among $u_3,u_4,u_5$, and applying it again gives a relation among $u_4,u_5,u_6$ — two equations in the two unknowns. Solving that pair pins down $u_5$. Tool #3 (Eliminate Possibilities) is the safety net: the answer must be one of the five listed values, and the algebra lands on exactly one of them.
Execute — Answer: B
6.EE.B.6 Step 1 Name the hidden terms
- Between the known terms $u_3=9$ and $u_6=128$ sit two unknown terms, $u_4$ and $u_5$.
- Give them short names: let $a=u_4$ and $b=u_5$.
- The value we want is $b$.
💡 You can't compute the missing terms yet, so give them letters and let the rule tell you what they must be.
8.EE.C.8 Step 2 First equation from the rule
- Apply the rule $u_{n+2}=2u_{n+1}+u_n$ with $n=3$.
- That makes the newest term $u_5$, so $u_5=2u_4+u_3$.
- Substituting the names gives $b=2a+9$.
💡 Plugging one starting index into the rule turns the abstract recurrence into a concrete equation linking the two unknowns.
8.EE.C.8 Step 3 Second equation from the rule
- Apply the same rule with $n=4$, so the newest term is $u_6$: $u_6=2u_5+u_4$.
- Since $u_6=128$, substituting the names gives $128=2b+a$.
💡 Sliding the rule one step forward uses the known end value $u_6$ and gives a second, independent equation.
8.EE.C.7 Step 4 Solve the two equations
- From the first equation, $a=\tfrac{b-9}{2}$.
- Put that into $128=2b+a$: $128=2b+\tfrac{b-9}{2}$.
- Multiply through by $2$ to clear the fraction: $256=4b+(b-9)=5b-9$.
- So $5b=265$ and $b=53$.
- Since $b=u_5$, we get $u_5=53$, which is choice (B).
💡 Two equations in two unknowns pin down a single value; substituting one into the other collapses it to one linear equation for $b$.
6.EE.B.6 Between the known terms $u_3=9$ and $u_6=128$ sit two unknown terms, $u_4$ and $ 8.EE.C.8 Apply the rule $u_{n+2}=2u_{n+1}+u_n$ with $n=3$. That makes the newest term $u_ 8.EE.C.8 Apply the same rule with $n=4$, so the newest term is $u_6$: $u_6=2u_5+u_4$. Sin 8.EE.C.7 From the first equation, $a=\tfrac{b-9}{2}$. Put that into $128=2b+a$: $128=2b+\ Review
Reasonableness: Check the whole chain. With $u_5=53$, the first equation gives $u_4=\tfrac{53-9}{2}=22$. Then the rule rebuilds the sequence: $u_5=2\cdot22+9=53$ and $u_6=2\cdot53+22=128$, which matches the given $u_6$ exactly. The terms $9,22,53,128$ also grow in the roughly-tripling way the rule forces, so $53$ sitting between $22$ and $128$ is the right size. Among the choices $40,53,68,88,104$, only $53$ makes the rule reproduce $u_6=128$.
Alternative: Eliminate instead of substitute: keep $b=2a+9$ and $a+2b=128$. From the first, $b-2a=9$; add twice the first to the second's rearrangement — or just note $a=128-2b$ and plug into $b=2a+9$ to get $b=2(128-2b)+9=265-4b$, so $5b=265$ and $b=53$. Same value, reached by clearing $a$ directly.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the two unknown middle terms $a=u_4$ and $b=u_5$ so the recurrence can be written as equations.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Turning the recurrence at $n=3$ and $n=4$ into the two equations $b=2a+9$ and $128=2b+a$.)8.EE.C.7Solve linear equations in one variable (Substituting to collapse the system into $256=5b-9$ and solving for $b=53$.)
⭐ When terms are missing between two you know, name them with letters, write the rule twice to get two equations, and solve.
⭐ When terms are missing between two you know, name them with letters, write the rule twice to get two equations, and solve.
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