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A=2^100-2^99-....-2^2-2-1
=>2A=2^101-2^100-....-2^3-2^2-2
=>2A-A=2^101-2^100-...-2^3-2^2-2-2^100+2^99+...+2^2+2+1
=>2A-A=2^101-2^101+1=1
vậy A=1
=> A=2100- (299+298+........+22+2+1) (3)
B
Ta có: B=299+298+..........+22+2+1 (1)
=> 2B=2100+299+..........+23+22+2 (2)
Lấy (2) - (1) ta có: 2B-B=(2100+299+...........+23+22+2)-(299+298+.............+22+2+1)
=> B= 2100-1
Thay vào (3) ta có: A=2100- (2100-1)
=> A=2100-2100+1
=> A=1
Đặt \(A=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2\)
\(=\left(2^{100}-2^{99}\right)+\left(2^{98}-2^{97}\right)+...+\left(2^4-2^3\right)+4\)
\(=2^{99}\left(2-1\right)+2^{97}\left(2-1\right)+...+2^3\left(2-1\right)+4\)
\(=\left(2^3+2^5+...+2^{97}+2^{99}\right)+4\)
Đặt \(B=2^3+2^5+...+2^{97}+2^{99}\)
\(\Rightarrow4B=2^5+2^7+...+2^{99}+2^{101}\)
\(\Rightarrow4B-B=3B=2^{101}-2^3\)
\(B=\frac{2^{101}-2^3}{3}\)
\(A=B+4\)
\(=\frac{2^{101}-2^3}{3}+4\)
\(=\frac{2^{101}-8+12}{3}\)
\(=\frac{2^{101}+4}{3}\)
A= 2+22 +23+...........+298+299+2100
= (2+22+23+24+25) +.............+(296+297+298+299+230)
= 62 +.................+2(2+22+23+24+25)
= 62+...................+62
=>A CHIA HẾT CHO 62(ĐPCM)
A = 2 + 22 + 23 + 24 + ... + 298 + 299 + 2100
= (2 + 22 + 23 + 24 + 25) + .... + (296 + 297 + 298 + 299 + 2100)
= 62 + ... + 295.(2 + 22 + 23 + 24 + 25)
= 62 + ... + 295 . 62
= 62.(1 + ... + 295) \(⋮\)62