Question

Σ 8 over n=1 2^n-1 ???

Answer 1

hmmm we can approach this summation as in we can the sum of a finite geometric sequence, so

`\displaystyle\sum_(n=1)^(8)~2^(n-1)\implies \sum_(n=1)^(8)~1\cdot 2^(n-1) \\\\[-0.35em] ~\dotfill\\\\ \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_(i=1)^(n)\ a_1\cdot r^(i-1)\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\[-0.5em] \hrulefill\\ n=8\\ a_1=1\\ r=2 \end{cases}`

`\displaystyle\sum_(n=1)^(8)~1\cdot 2^(n-1)\implies S_8=1\left( \cfrac{1-2^8}{1-2} \right)\implies S_8=1\left(\cfrac{-255}{-1} \right)\implies S_8=255`