besides the obvious add 2? I need one using the term # in the formula.
2007-04-27 09:07:58 · 10 answers · asked by hallofamer20 2 in Science & Mathematics ➔ Mathematics
a_n = 2^n.
(The nth term is 2 raised to the nth power.)
2007-04-27 09:11:56 · answer #1 · answered by Anonymous · 2⤊ 0⤋
Each number is doubled.
2+2 =4
4+4=8
8+8=16
16+16=32
2007-04-27 16:11:42 · answer #2 · answered by tazzlair 2 · 0⤊ 0⤋
2 to the n power
2^2=4
2^3=8
2^4=16
etc.
2007-04-27 16:19:02 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2007-04-27 16:12:16 · answer #4 · answered by danjlil_43515 4 · 0⤊ 1⤋
multiply by 2 each number
2007-04-27 16:14:47 · answer #5 · answered by crwswimmer 2 · 0⤊ 0⤋
2^(n+1)
You start off with 2 in the power of zero, and then you keep adding one for each step.
2007-04-27 16:20:42 · answer #6 · answered by Sparks 6 · 1⤊ 1⤋
it's not add 2, it's multiply by 2:
a[n] = 2^n.
2007-04-27 16:13:55 · answer #7 · answered by Philo 7 · 3⤊ 0⤋
To get the next number, double the number before it.
or, where x is a number in the series.
x = (x -1)^2
2007-04-27 16:13:44 · answer #8 · answered by DanE 7 · 0⤊ 2⤋
multiply every number by two to get the next number.
2007-04-27 16:32:48 · answer #9 · answered by Chatterbox 3 · 0⤊ 0⤋
The formula is y=2x. Duh!
2007-04-27 16:22:03 · answer #10 · answered by yodummy200 1 · 0⤊ 1⤋