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:

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