Find the first four terms of the recursive sequence defined below.

Question

aBASE0= -2

aBASEn+1= (aBASEn)^2-4

Answer 1

This recursive sequence is defined as follows:

  a₀ = -2      [the first term]

  a_n+1 = a_n² - 4     [for all other terms, n≥0; note that you cannot find a₀ using this line]

 

Note that this is not "BASE" for a number base, but it is "SUB" for subscript, indicating the term.  Sometimes, sequences are written with the first term being a₁ and sometimes sequences are written with the first term being a₀.  This is because there are situations that make one or the other more convenient (for example, starting with elapsed time t=0 makes sense).

 

To find the value of later terms, a recursive definition requires that you find the value of the previous term, which requires that you find the value of the previous term, which requires that you find the value of the previous term, which requires that you find the value of the previous term, ... which requires that use the value of the first term.

 

That's why (joke) that the dictionary definition of "recursive" is "see recursive."

 

So,

    a₀ = -2

    a₁ = (a₀)² - 4  =  4 - 4  =  0

    a₂ = (a₁)² - 4 = ((a₀)²-4)² - 4 =...=  0² - 4  = -4 

        For a_n+1, you may use the previous term (a_n) if you have just calculated it, rather than calculating it recursively again

    a₃ = (-4)² - 4 = 12

 

The first four terms in the sequence are -2, 0, -4, 12.

Answer 2

a₀ = -2

 

a_{n + 1} = (a_n)² - 4

 

This means that each term, n + 1, is equal to the preceding term, n, squared

and then subtract five from that number

 

Given:   a₀ = -2

 

Then:    a₁ = (-2)² -4 = 0

 

            a₂ = (0)² - 4 = -4

 

            a₃ = (-4)² -4 =12

 

            a₄ = (12)² - 4 = 140

Answer 3

If I understand your notation

a₀=-2

a₁=(-2)² -4=4-4=0

a₂=(0)² -4=0-4=-4

a₃=(-4)² -4=16-4=12

a₄=(12)² -4=144-4=140