Question

Find the next two terms 1,5,17,53,161

Answer 1

Let
`a_n` denote the
`n`-th term in the sequence. By checking the forward differences, we observe

`a_2 - a_1 = 5 - 1 = 4 = 4\cdot3^0`

`a_3 - a_2 = 17 - 5 = 12 = 4\cdot3^1`

`a_4 - a_3 = 53 - 17 = 36 = 4\cdot3^2`

`a_5 - a_4 = 161 - 53 = 108 = 4\cdot3^3`

The pattern is

`\begin{cases} a_1 = 1 \\ a_n = a_(n-1) + 4\cdot3^(n-2) & \text{for } n\ge1 \end{cases}`

So we find

`a_6 = a_5 + 4\cdot3^(6-2) = 161 + 4\cdot3^4 = \boxed{485}`

`a_7 = a_6 + 4\cdot3^(7-2) = 485 + 4\cdot3^5 = \boxed{1457}`

We also could have solved for
`a_n` first. By substitution,

`a_(n-1) = a_(n-2) + 4\cdot3^(n-3) \\\\ ~~~~ \implies a_n = a_(n-2) + 4 \left(3^(n-2) + 3^(n-3)\right)`

`a_(n-2) = a_(n-3) + 4\cdot3^(n-4) \\\\ ~~~~ \implies a_n = a_(n-3) + 4 \left(3^(n-2) + 3^(n-3) + 3^(n-4)\right)`

`a_(n-3) = a_(n-4) + 4\cdot3^(n-5) \\\\ ~~~~ \implies a_n = a_(n-4) + 4 \left(3^(n-2) + 3^(n-3) + 3^(n-4) + 3^(n-5)\right)`

and so on. After so many iterations of this, we see the pattern

`a_n = a_(n-k) + 4 \displaystyle \sum_(\ell=1)^k 3^(n-2-(\ell-1)) = a_(n-k) + 4 \cdot 3^(n-1) \sum_(\ell=1)^k \frac1{3^\ell}`

so that for
`k=n-1`, we get

`\displaystyle a_n = a_1 + 4 \cdot 3^(n-1) \sum_(\ell=1)^(n-1) \frac1{3^\ell}`

Let
`S_(n-1)` be the remaining sum. We have

`S_(n-1) = \frac13 + \frac1{3^2} + \frac1{3^3} + \cdots + \frac1{3^(n-1)}`

`\frac13 S_(n-1) = \frac1{3^2} + \frac1{3^3} + \frac1{3^4} + \cdots + \frac1{3^n}`

`\implies \frac23 S_(n-1) = \frac13 - \frac1{3^n}`

`\implies S_(n-1) = \frac12 \left(1 - \frac1{3^(n-1)}\right)`

and so

`\displaystyle a_n = 1 + 4 \cdot 3^(n-1) \cdot \frac12 \left(1 - \frac1{3^(n-1)}\right) \\\\ a_n = 2\cdot3^(n-1) - 1`

Then

`a_6 = 2\cdot3^(6-1)-1 = 485`

`a_7 = 2\cdot3^(7-1)-1 = 1457`