Question

Aflați x e n dacă 1+3+5+7+...+x=4 la puterea 2019

Vă rog! Dau coroană! ​

Answer 1

$1+3+7+...+x = 4^{2019} \\ \\ \Big(\dfrac{x+1}{2}\Big)^{2} = 4^{2019} \\ \\ \Big(\dfrac{x+1}{2}\Big)^2 = 2^{2\cdot 2019} \\ \\ \Big(\dfrac{x+1}{2}\Big)^{2} = \Big(2^{2019}\Big)^2\\ \\ \dfrac{x+1}{2} = 2^{2019}\Rightarrow x+1 = 2^{2020}\Rightarrow x = 2^{2020}-1$