Question

How do you write the first six terms of the sequence `a_n=n^2+3`?

Answer 1

`4, 7, 12, 19, 28, 39`

Explanation:

Method 1 - Direct substitution

`a_1 = 1^2+3 = 1+3 = 4`

`a_2 = 2^2+3 = 4+3 = 7`

`a_3 = 3^2+3 = 9+3 = 12`

`a_4 = 4^2+3 = 16+3 = 19`

`a_5 = 5^2+3 = 25+3 = 28`

`a_6 = 6^2+3 = 36+3 = 39`

`color(white)()`
Method 2 - Differences

Since the formula is a quadratic one, its coefficients will be determined by the first `3` terms.

Use direct substitution to write down the first `3` terms:

`4color(white)(00000)7color(white)(0000)12`

Under the gaps between the terms write down the sequence of differences:

`4color(white)(00000)7color(white)(0000)12`
`color(white)(000)3color(white)(00000)5`

Under the gap between the two terms, write the difference:

`4color(white)(00000)7color(white)(0000)12`
`color(white)(000)3color(white)(00000)5`
`color(white)(000000)2`

To this last line, add as many copies of the final difference as you would like extra terms of the original sequence:

`4color(white)(00000)7color(white)(0000)12`
`color(white)(000)3color(white)(00000)5`
`color(white)(000000)2color(white)(00000)color(red)2color(white)(00000)color(red)2color(white)(00000)color(red)(2)`

Fill in extra terms on the line above by adding the differences:

`4color(white)(00000)7color(white)(0000)12`
`color(white)(000)3color(white)(00000)5color(white)(00000)color(red)(7)color(white)(00000)color(red)(9)color(white)(0000)color(red)(11)`
`color(white)(000000)2color(white)(00000)color(red)(2)color(white)(00000)color(red)(2)color(white)(00000)color(red)(2)`

Fill in extra terms on the line above by adding the differences:

`4color(white)(00000)7color(white)(0000)12color(white)(0000)color(red)(19)color(white)(0000)color(red)(28)color(white)(0000)color(red)(39)`
`color(white)(000)3color(white)(00000)5color(white)(00000)color(red)(7)color(white)(00000)color(red)(9)color(white)(0000)color(red)(11)`
`color(white)(000000)2color(white)(00000)color(red)(2)color(white)(00000)color(red)(2)color(white)(00000)color(red)(2)`