Question

Given the following sequences, find the first 4 terms:a_n = 6n + 10 f(n) = 4(3)^n-1

Answer 1

a) We have the sequence:

`a_n=6n+10`

The first four terms, starting with a_0 and up to a_3 are:

`\begin{gathered} a_0=6\cdot0+10=10 \\ a_1=6\cdot1+10=16 \\ a_2=6\cdot2+10=22 \\ a_3=6\cdot3+10=28 \end{gathered}`

b) The second sequence is:

`f(n)=4\cdot3^(n-1)`

Then, the first four terms are:

`\begin{gathered} f(0)=4\cdot3^(0-1)=4\cdot3^{-1_{}}=(4)/(3) \\ f(1)=4\cdot3^(1-1)=4\cdot3^0=4 \\ f(2)=4\cdot3^(2-1)=4\cdot3=12 \\ f(3)=4\cdot3^(3-1)=4\cdot3^2=4\cdot9=36 \end{gathered}`