Question

The recursive sequence is defined by the formula `t_n=2t_(n-1)+3`; and `t_1=-2`, how do you find `t_6`?

Answer 1

Find `t_n = 2^(n-1)-3` and hence `t_6 = 2^5-3 = 29`

Explanation:

Note that as `n` and `t_n` get larger, the sequence will approximate a geometric sequence with common ratio `2`.

So we can look for a general formula for terms of the form:

`t_n = 2^na + b`

for some constants `a` and `b`.

Then we find:

`2^na + b = t_n = 2t_(n-1)+3 = 2(2^(n-1)a + b) = 2^na+2b+3`

Hence we find `b = -3`

Then:

`-2 = t_1 = 2^1 a - 3 =2a-3`

Hence we find `a = 1/2`

So the general formula of a term of our sequence is:

`t_n = 2^(n-1)-3`

If you like, we can double check this formula:

`2t_(n-1)+3 = 2(2^(n-2)-3)+3 = 2^(n-1)-6+3 = 2^(n-1)-3 = t_n`

In particular:

`t_6 = 2^5-3 = 32-3 = 29`

Answer 2

`t_6=2^5-3`

Explanation:

Proposing `t_n = c_0 a^n+b` and substituting we get
`c_0a^n+b = 2(c_0a^{n-1}+b)+3` or
`c_0a^n-2c_0a^{n-1}=2b-b+3`.

Making `a = 2` and `b = -3` the equality is observed. So we get

`t_n = c_0 2^n-3` and also
`t_1=c_0 2^1-3->c_0=1/2`. Putting all together
`t_n=2^{n-1}-3->t_6=2^5-3`