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

##### 1 Answer

#### 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#

**Method 2 - Differences**

Since the formula is a quadratic one, its coefficients will be determined by the first

Use direct substitution to write down the first

#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)#