[ Geometric progression ]

Find the sum $6+66+666+\ldots+666 . .6$, where the last number has p sixes in its notation.
The number written using $\mathrm{k}$ sixes is equal to 2(10k-1)/3.

## Solution

The number written using $\mathrm{k}$ nines is equal to ( $10^{\mathrm{k}}-1$ ), therefore, the number written using $\mathrm{k}$ sixes is $6\left(10^{\mathrm{k}}-1\right) / 9=2\left(10^{\mathrm{k}}-1\right) / 3$. Therefore, the sum $6+66+666+\ldots+666 . .6$ is equal to $2 / 3\left((10-1)+\left(10^{2}-1\right)+\ldots+\left(10^{\mathrm{n}}-1\right)\right)$ $=2 / 3\left(\left(10+10^{2}+\ldots+10^{\mathrm{n}}\right)\right.$-n). Inside the parentheses, we have a geometric progression. Its sum is $10+10^{2}+\ldots+10^{\mathrm{n}}=\left(10^{\mathrm{n}+1}-10\right) / 9$. Therefore, we finally get that the original sum is equal to $2\left(10^{\mathrm{n}+1}-10\right) / 27-2 \mathrm{n} / 3$.

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