Precalculus Examples

$1$ , $9$ , $25$ , $49$ , $81$

Step 1

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

$8, 16, 24, 32$

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 $a_{n} = a n^{2} + b n + c$ .

$8$

Step 3

Solve for $a$ by setting $2 a$ equal to the constant second level difference $8$ .

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Step 3.1

Set $2 a$ equal to the constant second level difference $8$ .

$2 a = 8$

Step 3.2

Divide each term in $2 a = 8$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 a = 8$ by $2$ .

$\frac{2 a}{2} = \frac{8}{2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 a}{2} = \frac{8}{2}$

Step 3.2.2.1.2

Divide $a$ by $1$ .

$a = \frac{8}{2}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $8$ by $2$ .

$a = 4$

Step 4

Solve for $b$ by setting $3 a + b$ equal to the first first level difference $8$ .

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Step 4.1

Set $3 a + b$ equal to the first first level difference $8$ .

$3 a + b = 8$

Step 4.2

Substitute $4$ for $a$ .

$3 \cdot 4 + b = 8$

Step 4.3

Multiply $3$ by $4$ .

$12 + b = 8$

Step 4.4

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

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Step 4.4.1

Subtract $12$ from both sides of the equation.

$b = 8 - 12$

Step 4.4.2

Subtract $12$ from $8$ .

$b = - 4$

Step 5

Solve for $c$ by setting $a + b + c$ equal to the first term in the sequence $1$ .

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Step 5.1

Set $a + b + c$ equal to the first term in the sequence $1$ .

$a + b + c = 1$

Step 5.2

Substitute $4$ for $a$ and $- 4$ for $b$ .

$4 - 4 + c = 1$

Step 5.3

Simplify $4 - 4 + c$ .

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Step 5.3.1

Subtract $4$ from $4$ .

$0 + c = 1$

Step 5.3.2

Add $0$ and $c$ .

$c = 1$

Step 6

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

$a_{n} = 4 n^{2} - 4 n + 1$