Finding $S_n$ in terms of n for the sequence (6 + 66 + 666 + 6666 ...)

Question

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

Answer 1

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?

Answer 2

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.

Answer 3

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: $$ \frac23\cdot10^1+\frac23\cdot10^2+\cdots+\frac23\cdot10^n=\frac{20}3\frac{10^n-1}{10-1} $$

Answer 4

When I go way back to when I first learned and thought of this and the concepts that clicked back before I got so familiar and jaded to things I started explaining them as though they were totally obvious, I'd have thought of it like this:

$S_n= 6 + 66 + 666 + 6666 + ....... + \underbrace{6666.....6}_n =$

$6\times (1 + 11 + 111 + 1111 + ....... + \underbrace{11111......1}_n)=$

$6\times (9 + 99 + 999 + 9999+ ...... + \underbrace{99999..... 9}_n)\times \frac 19$

Now $6\times M \times \frac 19 = \frac 69 \cdot M = \frac 23 M$ so....

$= \frac 23(9 + 99 + 999 + 9999+ ...... + \underbrace{99999..... 9}_n)=$

$\frac 23([10 -1] + [100-1] + [1000-1] + [10000-1]+ ..... + [1\underbrace{0000....0}_n-1])=$

$\frac 23([10^1 -1] + [10^2-1] + [10^3-1] + [10^4 -1] + ...... + [10^n-1])=$

$\frac 23([10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] -[\underbrace{1+1 +1 +1 +....... + 1}_n])=$

$\frac 23([10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] -n)=$

$\frac 23\times [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] - \frac 23\times n$

Now this will feel like we are going a little backwards but we need to figure out what $ [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n]$ equals. So let's do that:

$ [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] =$

$(1\underbrace{0}_1 + 1\underbrace{00}_2 + 1\underbrace{000}_3 + 1\underbrace{0000}_4 + ..... + 1\underbrace{000...0}_n) =$

$\underbrace{1111.......1}_n0 =$

$10\times \underbrace{1111.......1}_n$

Now we repeat the earlier trick.

$10\times \underbrace{99999.......9}_n \times \frac 19=$

$10\times (\underbrace{10^n-1}_n)\times\frac 19$

(Hey, notice I skipped the step $99999.....9 = 100000.....0 -1$ and just skipped to $10^n-1$. This is because we did it before and like true mathematicians we think if we did it once it is automatically obvious and doesn't need to be explained twice.)

$=\frac {10}9(10^n-1)$.

So we go back and plug that in. Where were we? Oh, yeah:

$S_n = $

$\frac 23\times [10^1 + 10^2 + 10^3 + 10^4 + ...... +10^n] - \frac 23\times n=$

$\frac 23\times [\frac {10}9(10^n-1)] -\frac 23\times n$

$\frac {20}{27}(10^n-1) - \frac 23n$.

That's it we are done.

.....

The thing is, the person who wrote the solution assumed that $\underbrace{mmmmm....m}_n = m\times \frac {10^n-1}9=\frac m9(10^n-1)$ and $\frac 69$ immediately reduces to $\frac 23$ goes without saying.

While I'm on the subject, very closely related, you will probably see in the near future assuming:

$1+a+a^2 + a^3 + a^4 + ........ +a^{n-1} = \frac {a^n-1}{a-1}$

is completely obvious and well known. This follows for

$(a-1)(1+a+a^2 + a^3 + a^4 + ........ +a^{n-1}) = $

$(a + a^2 + a^3 + a^4 + a^5 + ....... + a^n) -(1+a+a^2 + a^3 + a^4 + ........ +a^{n-1})=$

$a^n-1$

so $(1+a+a^2 + a^3 + a^4 + ........ +a^{n-1})=\frac {a^n-1}{a-1}$.

This bit about $1111111....1 = \frac 19\times 99999.....1=\frac 19(10^n -1)$ is actually just an specific application of that because:

$11111.....1 = $

$1 + 10 + 100 + 1000 + 10000 + ....... +1\underbrace{0000....0}_{n-1} = $

$1 + 10 + 10^2 + 10^3 + 10^4 + ...... + 10^{n-1} =$

$\frac {10^n -1}{10 -1} = \frac {10^n-1}9 =\frac 19(10^n-1)$.

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Now of course if we knew $\underbrace{mmmmm.....m}_n = \frac m9(10^n-1)$ and that $111111 = 1+10 + 10^2 + ... + 10^{n-1} = \frac 19(10^n-1)$ so well that it was instinctual and obvious we could do this a lot faster.

$S_n = 6 + 66 +.... + 66...66 =$

$6[\frac 19([10-1] + [10^2-1] + .... + [10^n-1])]=$

$\frac 23([10+10^2 + .... + 10^n]-n)=$

$\frac 23(10\times {10^n-1}9 - n)=$

$\frac {20}{27}(10^n-1) - \frac 23n$