I can't seem to find the nth term for this sequence of numbers:
2, 6, 18, 54
2007-08-12 13:37:07 · 8 answers · asked by Miss ??? 3 in Science & Mathematics ➔ Mathematics
T[n] = 2*(3^(n-1))
2007-08-12 13:40:15 · answer #1 · answered by gudspeling 7 · 1⤊ 0⤋
Y(n)=2*3^(n-1)
2007-08-12 20:44:00 · answer #2 · answered by bradlelf 2 · 0⤊ 0⤋
Of course we can take a(n) = 2*3^(n-1).
However, we should notice that there are infinitely many expressions which give 2, 6, 18, 54 for the first four terms. Now consider b(n) = 2*3^(n-1) + 5*(n-1)*(n-2)*(n-3)*(n-4). This also has the first four terms as stated, but the fifth term here is 282, not 162. Notice that the 5 in the formula for b(n) is completely arbitrary. It could be anything, and the first four terms would still be 2, 6, 18, 54.
This is not a question in mathematics, it is a question in clairvoyance: Try to guess what the poser had in mind.
2007-08-13 11:49:04 · answer #3 · answered by Tony 7 · 0⤊ 1⤋
the answer is 2*(3)^(n-1)
such that if n = 1 2*3^0=2
if n=2 2*3^1=6
if n=3 2*3^2 =18
if n =4 2*3^3=54
2007-08-12 20:41:48 · answer #4 · answered by Stephen D 2 · 1⤊ 0⤋
every new term is three times the previous term so what is 54x3= ? is your answer.
for a general expression to find the nth term then:
term sub n = 2*(3)^(n-1)
2007-08-12 20:45:07 · answer #5 · answered by 037 G 6 · 0⤊ 0⤋
It looks like they are multiplying by 3 to get the next number.
2007-08-12 20:40:59 · answer #6 · answered by Paladin 7 · 0⤊ 0⤋
multiply by 3
2007-08-12 20:40:19 · answer #7 · answered by Anonymous · 0⤊ 0⤋
i can only see that the ratio is 3 times.
2007-08-12 20:47:03 · answer #8 · answered by herbman76 2 · 0⤊ 0⤋