Answer:
A)[tex]\frac{1}{1024},\frac{1}{4096}[/tex]
B) [tex]\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.[/tex]
C) [tex]\\a_{n}=nq^{n-1} \:for\:n=1,2,3,4,...[/tex]
Step-by-step explanation:
1) Incomplete question. So completing the several terms:[tex]\left \{a_{n}\right \}_{n=1}^{\infty}=\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},... \right \}[/tex]
We can realize this a Geometric sequence, with the ratio equal to:
[tex]q=\frac{1}{4}[/tex]
A) To find the next two terms of this sequence, simply follow multiplying the 5th term by the ratio (q):
[tex]\frac{1}{256}*\mathbf{\frac{1}{4}}=\frac{1}{1024}\\\\\frac{1}{1024}*\mathbf{\frac{1}{4}}=\frac{1}{4096}\\\\\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},\mathbf{\frac{1}{1024},\frac{1}{4096}}\right \}[/tex]
B) To find a recurrence a relation, is to write it a function based on the last value. So that, the function relates to the last value.
[tex]\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.[/tex]
C) The explicit formula, is one valid for any value since we have the first one to find any term of the Geometric Sequence, therefore:
[tex]\\a_{n}=nq^{n-1} \:for\:n=1,2,3,4,...[/tex]
