Number Pattern Rules - ISEE Middle Level: Mathematics Achievement

Repeat the cycle $2,5$. The pattern is periodic with period $2$, alternating between $2$ and $5$ indefinitely.

$a_1=2$, and $a_n=3a_{n-1}$. This recursive definition starts with $2$ and multiplies each subsequent term by $3$, forming a geometric sequence.

Multiply by $\frac{1}{3}$ each term. The pattern is a geometric sequence with a common ratio of $\frac{1}{3}$, reducing each term accordingly.

$a_n = 5\cdot 2^{n-1}$. This formula defines the geometric sequence with first term $5$ and common ratio $2$ for the $n$th term.

$a_n = 64\cdot \left(\frac{1}{2}\right)^{n-1}$. This formula expresses the geometric sequence starting at $64$ with common ratio $\frac{1}{2}$ for the $n$th term.

Square the term number: $a_n=n^2$. The pattern consists of perfect squares, where each term is the square of its position $n$.

Cube the term number: $a_n=n^3$. The pattern consists of perfect cubes, where each term is the cube of its position $n$.

Add consecutive odd numbers: $+3,+5,+7,+9,\dots$. The differences between terms are consecutive odd numbers starting from $3$, building the sequence cumulatively.

$a_n = n^2 + 1$. This formula adds $1$ to the square of $n$ to produce each term in the sequence.

Add $1,2,3,4,\dots$ (triangular numbers). The pattern forms triangular numbers by cumulatively adding consecutive positive integers starting from $1$.

$a_n = \frac{n(n-1)}{2}$. This formula computes the $(n-1)$th triangular number, summing integers up to $n-1$.

Multiply by $-1$ each term. Multiplying by $-1$ alternates the sign of each term while keeping the absolute value constant at $3$.

$a_n = 3(-1)^{n-1}$. The formula uses powers of $-1$ to alternate signs, starting positive for odd $n$ and negative for even $n$.

$a_n = 3 + 4(n-1)$. This formula captures the arithmetic sequence starting at $3$ with a common difference of $4$ to find the $n$th term directly.

$a_n = 20 - 4(n-1)$. This formula represents the arithmetic sequence starting at $20$ with a common difference of $-4$ for the $n$th term.

$a_n = 1 + \frac{n(n-1)}{2}$. This formula adds $1$ to the $(n-1)$th triangular number to generate each term.

Square the term number plus $1$: $a_n=(n+1)^2$. Each term is the square of $(n+1)$, producing shifted perfect squares starting from $4$.

Subtract $3$ each term. The pattern is an arithmetic sequence with a common difference of $-3$, decreasing each term by $3$.

Add $4$ each term. The pattern is an arithmetic sequence with a common difference of $4$, increasing each term by that constant.

Products of consecutive integers: $a_n=n(n+1)$. Each term is the product of $n$ and $(n+1)$, generating twice the triangular numbers.

$a_n = 7 + 5(n-1)$. This arithmetic formula uses the first term $7$ and common difference $5$ to find the $n$th term.

Add $2,3,4,5,\dots$. The differences start from $2$ and increase by $1$ each time, forming triangular numbers beginning at $1$.

$a_1=1$, $a_2=1$, and $a_n=a_{n-1}+a_{n-2}$. This defines the Fibonacci sequence starting with two $1$s and using the sum of the previous two terms for each subsequent one.

Add $1,2,3,4,\dots$. The differences are consecutive positive integers starting from $1$, accumulating to form the sequence.