Question

Consider the following sequence. -1,2,10,23,41... complete the table table below for the sequence.

Answer 1

Answer:

• a) see the attachment for a table of values
• b) a[n] = 5 + 2·a[n-1] -a[n-2]
• c) this is a quadratic function

Step-by-step explanation:

a) Apparently, you're to put the given values into a table. The first 5 entries of the table below are the given values. The next few are the result of using the recursive formula. (The formula bar shows the formula that is in the selected cell.)

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b) The first differences of the terms of this sequence are ...

• 2 -(-1) = 3
• 10 -2 = 8
• 23 -10 = 13
• 41 -23 = 18

These are not constant, so the sequence is not arithmetic. The ratios of terms are not constant (2/-1 ≠ 10/2), so the sequence is not geometric.

The second differences are ...

• 8 - 3 = 5
• 13 - 8 = 5
• 18 - 13 = 5

These are constant, which tells us the sequence is a polynomial sequence of 2nd degree (since 2nd differences are constant).

In terms of the differences and second differences we can write the expression for the n-th term

  first difference with term before: a[n] -a[n-1]

  first difference between previous two terms: a[n-1] -a[n-2]

The difference between these two differences is 5, so we can write ...

  (a[n] -a[n-1]) -(a[n-1] -a[n-2]) = 5

Solving for a[n], we get ...

  a[n] = 5 + 2·a[n-1] -a[n-2] . . . . . the desired recursive relation

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c) As indicated in part (b), this sequence is quadratic. As "proof", we offer that the sequence can be described by an explicit quadratic formula that can be derived from the first sequence term (d0) and the first and second differences (d1 and d2):

  f(n) = d0 + (n-1)(d1 +(n-2)/2·d2) = -1 +(n-1)(3 +(n-2)/2·5)

  f(n) = 5/2n² -9/2n +1