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Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?


Stage: 4 and 5 Challenge Level: Challenge Level:1

The $n^{th}$ term of a sequence is given by the formula $n^3 + 11n$. Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by $6$.

Congratulations to Julia Collins, age 17, Langley Park School for Girls, Bromley; Kookhyun Lee; Yatir Halevi age 17; Sim S K age 14; Ang Zhi Ping age 16 for your splendid solutions.

This is Kookhyun Lee's solution: First term: $12$, second term: $30$, third term: $60$, fourth term: $108$. The $100$th term is $100^3 +1100 = 1001100$. The $99$th term is $970299 + 1089 = 971388$ so the first term bigger than a million is $1001100$ when $n=100.$

Proof that all the terms are divisible by $6$.

$$n^3 + 11n = n^3 + 12n - n = n(n^2-1) + 12n$$

This must be a multiple of 6 because $n(n^2-1)$ can be written as $(n-1)\times n \times (n+1)$. Any multiple of three consecutive integers is a multiple of $6$ because it contains a multiple of two (an even number) and a multiple of three.

The following solution, uses a different method. It arrived early in the morning on the first day that the question was published from Yatir Halevi, age 17.5, Maccabim and Reut High-School, Israel.

We have a sequence given by the formula $n^3+11n.$ We have to find the first value of $n$ for which $n^3+11n> 10^6$ and we could use

The second way is a much nicer one. We notice that $100^3$ is $10^6$, so we know for $n=100$ that $n^3+11n$ is bigger than $10^6$, so we check $n=99$ and we get: $971388$ which is smaller than $10^6$. So we have the answer: $n=100$ is the first $n$ for which $n^3+11n$ is bigger than $10^6$.

The next thing we have to prove is that $n^3+11n$ is always divisible by $6$. This we will prove by using modular arithmetic. We will use modulus $6$. For each $n$, we can have a residue of either: $0$, $1$, $2$, $3$, $4$ or $5$. For $n^3$ we get the following residues:

$0$, $1$, $2$, $3$, $4$, $5$ respectively (to $n$). For $11n$ we get the following residues: $0$, $5$, $4$, $3$, $2$, $1$ respectively (to $n$).

Combining $n^3$ and $11n$ (respectively) we get a $0$ residue, because: $0+0=0$ (mod $6$), $1+5=6=0$ (mod $6$), $2+4=6=0$ (mod $6$), $3+3=6=0$ (mod $6$), $4+2=6=0$ (mod $6$), $5+1=6=0$ (mod $6$). This means that we get a zero residue when dividing by $6$, or in other words, $(n^3+11n)$ is a multiple of $6$ or $6$ divides $n^3+11n$.