Question

2^x= 1+2^0+2^1+2^2+.....+2^1968

Answer 1

$S_n=2^{0} + 2^{1} + ...+ 2^{n} |*2 \Rightarrow 2S_n = 2^{1} + 2^{2} + ... + 2^{n+1} \\ \\ 2^{1} + 2^{2} + ... + 2^{n} = S_n -2^{0} = S_n-1 \\ \\ \Rightarrow 2S_n =S_n - 1 + 2^{n+1} \Rightarrow S_n = 2^{n+1} -1 (formula \ generala) \\ \\ \\ \\S_n = 2^{x}-1\\ \\ 2^{x} = 1 + 2^{0} + 2^{1} + ... + 2^{1968} \Rightarrow S_n = 2^{0} + 2^{1} + 2^{3} + ...+ 2^{1968} |*2 \Rightarrow 2S_n= 2^{1} +2^{2} + ...+ 2^{1969} \Rightarrow S_n = 2^{1969}-1 \\ \\ \\2^{x} - 1 = 2^{1969}-1 \Rightarrow x=1969$