Identify the Sequence 36 , 49 , 64 , 81

$36$ , $49$ , $64$ , $81$

Step 1

Find the first level differences by finding the differences between consecutive terms.

$13, 15, 17$

Step 2

Find the second level difference by finding the differences between the first level differences. Because the second level difference is constant, the sequence is quadratic and given by $a_{n} = a n^{2} + b n + c$ .

$2$

Step 3

Solve for $a$ by setting $2 a$ equal to the constant second level difference $2$ .

Set $2 a$ equal to the constant second level difference $2$ .

$2 a = 2$

Divide each term in $2 a = 2$ by $2$ and simplify.

Divide each term in $2 a = 2$ by $2$ .

$\frac{2 a}{2} = \frac{2}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{2}{2}$

Divide $a$ by $1$ .

$a = \frac{2}{2}$

Simplify the right side.

Divide $2$ by $2$ .

$a = 1$

Step 4

Solve for $b$ by setting $3 a + b$ equal to the first first level difference $13$ .

Set $3 a + b$ equal to the first first level difference $13$ .

$3 a + b = 13$

Substitute $1$ for $a$ .

$3 \cdot 1 + b = 13$

Multiply $3$ by $1$ .

$3 + b = 13$

Move all terms not containing $b$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$b = 13 - 3$

Subtract $3$ from $13$ .

$b = 10$

Step 5

Solve for $c$ by setting $a + b + c$ equal to the first term in the sequence $36$ .

Set $a + b + c$ equal to the first term in the sequence $36$ .

$a + b + c = 36$

Substitute $1$ for $a$ and $10$ for $b$ .

$1 + 10 + c = 36$

Add $1$ and $10$ .

$11 + c = 36$

Move all terms not containing $c$ to the right side of the equation.

Subtract $11$ from both sides of the equation.

$c = 36 - 11$

Subtract $11$ from $36$ .

$c = 25$

Step 6

Substitute the values of $a$ , $b$ , and $c$ into the quadratic sequence formula $a_{n} = a n^{2} + b n + c$ .

$a_{n} = 1 n^{2} + 10 n + 25$

Step 7

Multiply $n^{2}$ by $1$ .

$a_{n} = n^{2} + 10 n + 25$

Identify the Sequence 36 , 49 , 64 , 81

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