Question

Given `a_1=4`, what is the 7th term of the recursive sequence using `a_n= 2a_(n-1) - 3`, for `n>=2`?

Answer 1

`a(7)=67`

Explanation:

Let's look at the first few numbers in this sequence:
`a(1)=4`
`a(2)=2*a(1)-3=2*4-3=5`
`a(3)=2*a(2)-3=2*5-3=7`
`a(4)=2*a(3)-3=2*7-3=11`
`a(5)=2*a(4)-3=2*11-3=19`

Now, let's look at the difference between them at each step
`a(2)-a(1)=5-4=1`
`a(3)-a(2)=7-5=2`
`a(4)-a(3)=11-7=4`
`a(5)-a(4)=19-11=8`

That sequence of numbers (1,2,4,8,...) is the sequence of the powers of two (`2^0, 2^1, 2^2, 2^3,...)`. Also notice that the number in the sequence that we are subtracting from is 2 more than the power of two we get from the subtraction (i.e. the `0` in `2^0` is two less than the `n` in a(2)). That allows us to write:

`color(red)(a(n)=4+2^(n-2)`
`color(red)"for any"` `color(red)(n >=2)`

Now, that is fine and dandy, and will give you the answer for any number in the sequence but I (and math teachers in general) like elegant answers and I think we can come up with a better equation for the sequence.

Let's look at the sequence of powers of two next to the this sequence member by member:

`2^0=1 rarr a(1)=4`
`2^1=2 rarr a(2)=5`
`2^2=4 rarr a(3)=7`
`2^3=8 rarr a(4)=11`
`2^4=16 rarr a(5)=19`

Notice something interesting? Each number in our sequence is the correspondent power of two plus 3. Now we can write that:

`color(red)(a(n)=2^(n-1)+3`
`color(red)"for any n"` (as an added elegant bonus it is a nice mirror of the original equation)

So `a(7)=2^6+3=64+3=67`