Question

Σ 10 over n=1 8(2^n-1) = [?]

Answer 1

`\displaystyle\sum_(n=1)^(10)~8(2^(n-1))\implies \sum_(n=1)^(10)~8(2^n\cdot 2^(-1))\implies \sum_(n=1)^(10)~8\cdot 2^n\cdot \cfrac{1}{2}\implies \sum_(n=1)^(10)~4\cdot 2^n \\\\\\ \displaystyle 4\sum_(n=1)^(10)~2^n\implies 4\cdot \begin{array}{llll} 2^1&2\\ 2^2&4\\ 2^3&8\\ 2^4&16\\ 2^5&32\\ 2^6&64\\ 2^7&128\\ 2^8&256\\ 2^9&512\\ 2^(10)&1024\\\cline{1-2} &2046 \end{array}\implies 4(2046)\implies 8184`