Let a = 111 . . . 1, where the digit 1 appears 2018 consecutive times. Let b = 222 . . . 2, where
in this case the digit 2 appears 1009 consecutive times. Without using a calculator, evaluate
√(a − b)
Here is what you are asked to do: $\sqrt {\sum\limits_{k = 0}^{2018} {{{10}^k}} - 2 \cdot \sum\limits_{k = 0}^{1009} {{{10}^k}} }$.
Now I have no idea what the point of this question may be. Nor can I imagine what it is designed to teach.
With a calculator see here.
Without, I think this probability busywork.
for all $n$
$\sum _{k=0}^{2n-1} 10^k-2\sum _{k=0}^{n-1} 10^k=\left(3\sum _{k=0}^{n-1} 10^k\right){}^2$
to prove this, note that each sum is a geometric series so for example
$\sum _{k=0}^{2n-1} 10^k=\frac{10^{2n}-1}{9}$
etc.