Finding nth term for a recursive/iterative/term to term sequence

Question

I have the sequence: 1, 16, 166, 1666 . Im trying to find the nth term for this sequence but since it is not linear or geometric I'm not sure how to. I worked out that the recursive formula is: \begin{gather} a_1 = 1 \\ a_{n + 1} = 10a_{n} + 6 \end{gather} I'm trying to write this as an nth term and I saw that one solution was: \begin{gather} \frac{1}{6}\left ( 10 ^{n}-4\right ) \end{gather} But I don't know how to derive this. I have seen other answers on this site but the explanations arent often very clear or involve math that is beyond my ability. For reference I am a high schooler in my final year, if anyone is familiar with the British Education system I am studying further maths for a level so I don't know any math more complex than that. Thank you

Answer 1

$a_n = 1\underbrace{6666...6}_{n-1}=$

$1\underbrace{0000.....0}_{n-1} + \underbrace{666...6}_{n-1} =$

$10^{n-1} + 6\cdot \underbrace{111....1}_{n-1} =$

$10^{n-1} + 6\cdot \frac {\underbrace{9999....9}_{n-1}}9 =$

$10^{n-1} + 6\cdot \frac {10^{n-1} -1}9=$

$10^{n-1} + \frac 23(10^{n-1} -1) =$

$\frac {3\cdot 10^{n-1} +2\cdot 10^{n-1} - 2}3=$

$\frac {5\cdot 10^{n-1} -2}3$.

......

ALteratively:

$a_n = 1\underbrace{6666...6}_{n-1}$

$3a_n = 4\underbrace{99999...9}_{n-2}8=$

$5\underbrace{00000.....0}_{n-1} - 2$

$5\cdot 10^{n-1} -2$

So $a_n = \frac {5\cdot 10^{n-1} -2}3$

Answer 2

Observe that $1/6= 0.1666\dots $ and $a_n = 10^n(0.1666\dots - \varepsilon_n)$, where $\varepsilon_n = 10^{-n}\cdot 0.666 \dots$. Since $\varepsilon_n$ is geometric for every $n$, you can have a closed expression which is $$\varepsilon_n = 10^{-n} \cdot \dfrac{4}6. $$ Hence, $$a_n = \dfrac{1}{6} \bigg(10^n - 4\bigg).$$

Answer 3

The Online Encyclopedia of Integer Sequences A246057 shows$\quad a(n) = (5\times10^n - 2)/3$

It generates $\\ 16\quad 166\quad 1666\quad 16666\quad 166666\quad 1666666\quad 16666666\quad 166666666\quad \cdots$

Working backwards, we see that all of these number resemble $x+\dfrac23$ so we can try $$3\times 16=48\qquad 3\times 166=498\qquad 3\times1666=4998 \quad\cdots\quad $$

and we can see that each is $2$ short of $5\times10^n.\quad$ From here it should be easy to see how the formula was developed and how it works.