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bài làm
2n - 1 - 2 - 22 - 23 - .............. - 2100 = 1
2n - ( 1 + 2 - 22 + 23 + ........ + 2100 ) = 1
2n - ( 2101 - 1 ) = 1
2n - 1 = 2101 - 1
=> 2n = 2101
Vậy n = 101
TA CO 2^n-1-2^2-....-2^100=1
=>2n-(1+2^2+2^3+...+2^100)=1
dat A=1+2^2+2^3+...+2^100
=>2A=2+2^3+2^4+...+2^101
=>2A-A=(2+2^2+2^3+...+2^101)-(1+2+2^2+...+2^100)
=>A=2+2^2+2^3+...+2^101-1-2-2^2-..-2^100
=>A=2^101-1
=>2^n-(2^101-1)=1
=>2^n-2^101+1=1
=>2^n=1-1+2^101
=>2^n=2^101=>n=101
Vay n=101.
2^n-1-2^2-....-2^100=1
=>2n-(1+2^2+2^3+...+2^100)=1
dat A=1+2^2+2^3+...+2^100
=>2A=2+2^3+2^4+...+2^101
=>2A-A=(2+2^2+2^3+...+2^101)-(1+2+2^2+...+2^100)
=>A=2+2^2+2^3+...+2^101-1-2-2^2-..-2^100
=>A=2^101-1
=>2^n-(2^101-1)=1
=>2^n-2^101+1=1
=>2^n=1-1+2^101
=>2^n=2^101=>n=101
Vay n=101.
Đặt B=2+2^2+2^3...+2^100
2B=2^2+2^3+2^4+.....+2^101
2B-B=2^2+2^3+...+2^101-2-2^2-...-2^100
B=2^101-2
Ta có:2^n-1-2-2^2-2^3-...-2^100=1
<=>2^n-1-(2+2^2+...+2^100)=1
<=>2^n-1-B=2^n-1-(2^101-2)=1
<=>2^n-(2^101-2)=2
<=>2^n=2+2^101-2=2^101
<=>n=101
Vậy n=101