Question

Sa se determine numarul natural ''n'' din egaliatea 1+2+4+8+...+2 la puterea n+1= 1023

Answer 1

1+2+4+8+...+2^(n+1)1023.  
1+2+2^2+2^3+...+2^(n+1)1023
2+2^2+2^3+2^4+...+2^(n+1)+2^(n+2)1023*2
1+2+2^2+2^3+...+2^(n+1)+2^(n+2)=1023*2+1
1023+2^(n+2)=1023*2+1
2^(n+2)=1023+1=1024
2^10=1024 rezulta ca n+2=10 , n=8
*2^2= doi la puterea a doua