Finding $S_n$ in terms of n for the sequence (6 + 66 + 666 + 6666 ...) Sequence given : 6, 66, 666, 6666. Find $S_n$ in terms of n
The common ratio of a geometric progression can be solved is $\frac{T_n}{T_{n-1}} = r$, where r is the common ratio and n is the
When plugging in 66 as $T_n$ and 6 as $T_{n-1}$, I got the following ratio: $ \frac {66}{6} = 11$.
However, when I plugged in 666 as $T_n$ and 66 as $T_{n-1}$, I got: $\frac {666}{66} = 10.09$.
And when I plugged in 6666 and 666: $ \frac {6666}{666} = 10.009$.
It's clear to me that the ratio is slowly decreasing, and seems to be approaching 10. Alas, this is about as far as I have gotten.
Looking at the answers scheme, the final answers is $ \frac {20}{27}{(10^n-1)} - \frac {2}{3}{(n)}.$
The answers scheme does include a few steps, but frankly, I couldn't understand the reasoning behind them, but I guess I should post them anyways.
$$ \frac {2}{3}[9 + 99 + 999 + 9999] $$
$$ S_n = \frac {2}{3}{(10-1)} + (10^2-1) + ...(10^n-1) $$
$$ = \frac {2}{3}[10^1+10^2+10^3...10^n] + \frac {2}{3}[-1-1-1-1...-1] $$
$$ = \frac {2}{3}(10)\left(\frac {10^n-1}{10-1}\right) - \frac {2}{3}(n) $$
$$ \frac {20}{27}{(10^n-1)} - \frac {2}{3}{(n)} $$
I'm sorry if I did something stupid, but I have no idea where that $\frac {2}{3}$ came from, and even if I do, I don't understand the reasoning or the explanation behind it. I already asked my teacher, as well as my mother, both of which yielded little in understanding the logic behind the solutions given in the answers scheme.
If anybody could offer an explanation, it would be greatly appreciated
 A: Let’s look at it another way:
$$6666=6\times 1111=6\times \dfrac{9999}{9}$$$$=\dfrac69\cdot(10^4-1)=\dfrac23(10^4-1).$$
Thus, the general term is $$T_n=\underbrace{6666…6}_{n \ \text{times}\ 6}=6\times\dfrac{\overbrace{9999…9}^{n\ \text{times}\ 9}}{9}$$$$=\dfrac69\times (10^n-1)=\dfrac23(10^n-1)$$
So $$S_n=\sum_{i=1}^n T_i= \sum_{i=1}^n \dfrac23(10^i-1) = \dfrac23 \sum_{i=1}^n(10^i-1)$$$$=\dfrac23\bigg(\sum_{i=1}^n(10^i)\bigg)-\frac23 n$$ Can you take it from here?
A: There are many ways, including saying $$T_n=\frac23(10^n-1)$$ and so $$S_n=\sum\limits_{k=1}^n T_k=\frac23\left(\sum\limits_{k=1}^n 10^k - \sum\limits_{k=1}^n 1 \right)$$ where the sums are simple geometric series.
Personally I prefer

*

*$S_n = 6+66+ 666+ \cdots + 66\cdots 6 + 66\cdots 66$

*$10S_n+6n = 66+666+ 6666+ \cdots + 66\cdots 66 + 66\cdots 666$

*$9S_n+6n = 10S_n+6n-S_n= 66\cdots 666 - 6$

*etc.

A: Alternately, note that if you add $\frac23=0.666\ldots$ to each term, then it actually becomes a geometric progression with common ratio $10$. So remembering to subtract it again afterwards, we get
$$
S_n=T_1+T_2+\cdots+T_n\\
=\frac23\cdot10^1+\frac23\cdot10^2+\cdots+\frac23\cdot10^n-n\cdot\frac23
$$
And now you can use the standard formula for sum of geometric series on the first part:
$$
\frac23\cdot10^1+\frac23\cdot10^2+\cdots+\frac23\cdot10^n=\frac{20}3\frac{10^n-1}{10-1}
$$
