Hello, Trenters4325!
Prove that: \(\displaystyle \,2^n\,-\,2^{n-1}\,-\,2^{n-2}\,-\,\cdots\,-\,2^0\;=\;1\;\;\)for \(\displaystyle n\,\geq\,1\)
We have: \(\displaystyle \,S\;=\;2^n\,-\,\underbrace{\left[1\,+\,2\,+\,2^2\,+\,\cdots\,+\,2^{n-1}\right]}\)
. . . . . . . . . . . . . . . . . . . . . . a geometric series
The series has first term \(\displaystyle a\,=\,1\), common ratio \(\displaystyle r\,=\,2\), and \(\displaystyle n\) terms.
Its sum is: \(\displaystyle \,1\cdot\frac{1\,-\,2^n}{1\,-\,2}\;=\;2^n\,-\,1\)
Therefore: \(\displaystyle \:S\;=\;2^n\, -\, \left(2^n\,-\,1)\;=\;1\)
