Question

Formula la suma asta : 2^1+2^2+2^3+.......2^2017=
Ofer coroana.

Answer 1

S = 2^1 + 2^2 + 2^3 + ... + 2^2017 /(*2) →

2S= 2^2 + 2^3 + ... + 2^2017 + 2^2018

2^2 + 2^3 + ... + 2^2017 = S - 2^1

2S = S-2 + 2^2018 / (-S) →

S = 2^2018 - 2

Answer 2

$\boxed{ 2^1+2^2+2^3+.......+2^n = 2^{n+1} - 2} \quad ( formula ) \\ \\ 2^1+2^2+2^3+.......2^{2017} = 2^{2017+1}-2 = 2^{2018}-2 \\ \\ Teoretic ar trebui demonstrata prin calcule aceasta formula, dar \\ consider ca este prea simpla ca sa nu fie folosita.$

$\ Demonstratie: \\ \\ S = 2^1+2^2+2^3+...+2^{2016}+2^{2017} \Big|\cdot 2\quad \quad ( inmultim cu 2 suma) \\ \\ 2\cdot S = 2^2+2^3+2^4+...+2^{2017}+2^{2018} \\ 2\cdot S = 2^1-2^1 +2^2+2^3+2^4+...+2^{2017}+2^{2018} \\ ( doar am adaugat un 2 si l-am scazut, nu am influentat deloc suma ) \\ \\ 2 \cdot S = 2^1+2^2+2^3+2^4+....+2^{2017}+2^{2018} -2^1 \\ (l-am mutat pe -2 la capat ca sa se observe ca, 2^1+2^2+...+2^{2017} \\ \ este chiar S, inlocuim acea suma cu S )$

$\\ \\ 2\cdot S = S + 2^{2018} - 2^1 \Rightarrow 2\cdot S - S = 2^{2018}-2^1 \Rightarrow \boxed{S = 2^{2018}-2}$