Respuesta :
Every next term appears to be three times the previous term, so the sequence is defined recursively by
[tex]a_n=3a_{n-1}[/tex]
and explicitly by
[tex]a_n=3a_{n-1}=3^2a_{n-2}=\cdots=3^{n-1}a_1[/tex]
Given that [tex]a_1=2[/tex], you have
[tex]a_n=2\times3^{n-1}\implies a_{15}=2\times3^{14}=9565938[/tex]
[tex]a_n=3a_{n-1}[/tex]
and explicitly by
[tex]a_n=3a_{n-1}=3^2a_{n-2}=\cdots=3^{n-1}a_1[/tex]
Given that [tex]a_1=2[/tex], you have
[tex]a_n=2\times3^{n-1}\implies a_{15}=2\times3^{14}=9565938[/tex]
Answer:
[tex]a_{15} = 2(3)^{14}[/tex]
Step-by-step explanation:
the sequence 2, 6, 18, 54, ...,
3 is multiplied with first term to get 6
2*3 = 6
6* 3= 18
18 * 3= 54
Each term is multiplied with 3 to get next term
Given sequence is geometric
To get nth term we use formula
[tex]a_n = a_1(r)^{n-1}[/tex]
Where 'a1' is the first term
r is the common ratio
n is the number of terms
first term a1= 2
Each term is multiplied with 3 to get next term, so r= 3
now we need to find out the fifteenth term so n= 15
[tex]a_{15} = 2(3)^{15-1}[/tex]
[tex]a_{15} = 2(3)^{14}[/tex]