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

A)
`(1)/(1024),(1)/(4096)`

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

C)
`\\a_(n)=nq^(n-1) \:for\:n=1,2,3,4,...`

Explanation:

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

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

`q=(1)/(4)`

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

`(1)/(256)*\mathbf{(1)/(4)}=(1)/(1024)\\\\(1)/(1024)*\mathbf{(1)/(4)}=(1)/(4096)\\\\\left \{ 1,(1)/(4),(1)/(16),(1)/(64),(1)/(256),\mathbf{(1)/(1024),(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)*(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,...`