Question

Several terms of a sequence {an}n=1 infinity are given. A. Find the next two terms of the sequence. B. Find a recurrence relation that generates the sequence. C. Find an explicit formula for the general nth term of the sequence.

Answer 1

Answer:

A)$\frac{1}{1024},\frac{1}{4096}$

B) $\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.$

C) $\\a_{n}=nq^{n-1} \:for\:n=1,2,3,4,...$

Step-by-step explanation:

1) Incomplete question. So completing the several terms:$\left \{a_{n}\right \}_{n=1}^{\infty}=\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},... \right \}$

We can realize this a Geometric sequence, with the ratio equal to:

$q=\frac{1}{4}$

A) To find the next two terms of this sequence, simply follow multiplying the 5th term by the ratio (q):

$\frac{1}{256}*\mathbf{\frac{1}{4}}=\frac{1}{1024}\\\\\frac{1}{1024}*\mathbf{\frac{1}{4}}=\frac{1}{4096}\\\\\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},\mathbf{\frac{1}{1024},\frac{1}{4096}}\right \}$

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.

$\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.$

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:

$\\a_{n}=nq^{n-1} \:for\:n=1,2,3,4,...$