<s>[INST] Number of digits from $1$ to $n$ Let $n$ be a natural number $k$ digits. Show that the quantity $Q$ of digits required to write the natural numbers from $1$ to $n$ is:
$Q = k(n+1) - \underbrace{111\ldots11}_{k\textrm{ digits}}$
 [/INST] As this question is now 14 hours old, I will give an answer.
First sum the total number of digits of all the natural numbers $ \le k-1.$ Call this sum $S.$
There are $9 \times 10^{i-1}$ natural numbers of length $i,$ so the total number of digits of  natural numbers of length $i$ is $9i \times 10^{i-1}.$ Hence
$$ S=9(1+2.10 + 3.10^2 + \cdots + (k-1)10^{k-2}) = (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}.$$
Now the first natural number with $k$ digits is $10^{k-1}$ and so there are $n- 10^{k-1}+1$ natural numbers from $10^{k-1}$ to $n$ inclusive. Each of these numbers has $k$ digits and so the number of digits required to write the natural numbers from $1$ to $n$ is
$$  k( n- 10^{k-1}+1) + S =  k( n- 10^{k-1}+1) + (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}$$
$$ = k(n+1) - \frac{10^k-1}{9}.$$
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