Question

NEED HELP WILL MEDAL AND FAN. what are the first four terms of the following sequence: An=2^n/3! for n (equivalent or equal to) 1.?

Answer 1

the sequence is \[\Large A_{n} = \frac{2^n}{3!}\] right?

Answer 2

yes

Answer 3

re-write the 3! on the bottom, I would think.

Answer 4

3! = 1 * 2 * 3

Answer 5

yes @jim_thompson5910

Answer 6

I was thinking to rew-rite it as:
\[a_n=\frac{1}{6}2^n\]

Answer 7

To find the first term \(\Large A_{1}\), replace n with 1 and evaluate
\[\Large A_{n} = \frac{2^n}{3!}\]
\[\Large A_{1} = \frac{2^1}{3!}\]
\[\Large A_{1} = \frac{2^1}{3*2*1}\]
\[\Large A_{1} = \frac{2}{6}\]
\[\Large A_{1} = \frac{1}{3}\]

Answer 8

and yes,

\[\Large A_{n} = \frac{2^n}{3!} = \frac{1}{6}*2^n\]

Answer 9

you will do the same for n = 2, n = 3, n = 4 to get terms 2 through 4

Answer 10

so first term is 1/3, second is 2/3, third is 4/3, and fourth is 8/3?

Answer 11

Or, \[a_n=\frac{1}{3}2^{n-1}\]

Answer 12

using which ever version of a_n, what is the second term (when n=2) ?

Answer 13

sar12389, that is correct

Answer 14

didn't notice (s)he posted them . lol

Answer 15

another question, Find The first four terms of the following sequence : An=2^n

Answer 16

a(1) (when n=1) is?

Answer 17

same idea, just different formula

Answer 18

i tired it and i got 2,4,6, and 16

Answer 19

2^3 is NOT equal to 6

Answer 20

you mixed that up with 2*3

Answer 21

i meant 8 instead of 6 @jim_thompson5910

Answer 22

good

Answer 23

2,4,8,16 is correct

Answer 24

thank you!

Answer 25

np