A = (2 + 22 + 23 + 24) + (25 + 26 + 27 + 28) + ... + (257 + 258 + 259 + 260)
= (2 + 22 + 23 + 24) + 24.(2 + 22 + 23 + 24) + ... + 256.(2 + 22 + 23 + 24)
= 30 + 24.30 + ... + 256.30
= 30."(1 + 24 + ... + 256)
= 5.6.(1 + 24 + ... + 256) \(⋮\)5
=> \(A⋮5\left(\text{đpcm}\right)\)
Ta có : A = 2 + 22 + 23 + ... + 260
2A = 22 + 23 + ... + 260 + 261
2A - A = 261 - 2
A = 261 - 2
Vì 261 - 2 = 24x15+1 - 2 = ( 24)15 x 2 - 2 = 1615 x 2 - 2 = ....6 x 2 - 2 = ....2 - 2 = ....0
Mà ....0 chia hết cho 5
261 - 2 chia hết cho 5
2 + 22 + 23 + ... + 260 chia hết cho 5 ( đpcm )
Vậy A chia hết cho 5
\(A=2+2^2+2^3+2^4+...+2^{59}+2^{60}\) có 60 số hạng.
\(A=\left(2+2^3+2^5+...+2^{59}\right)+\left(2^2+2^4+...+2^{60}\right)\)
có ( 59 - 1):2 +1 = 30 số hạng có: ( 60 - 2) : 2 + 1= 30 số hạng.
\(A=\left[\left(2+2^3\right)+\left(2^5+2^7\right)+...+\left(2^{57}+2^{59}\right)\right]+\left[\left(2^2+2^4\right)+\left(2^6+2^8\right)+...+\left(2^{58}+2^{60}\right)\right]\)
\(A=\left[2\left(1+2^2\right)+...+2^{57}\left(1+2^2\right)\right]+\left[2^2\left(1+2^2\right)+...+2^{58}\left(1+2^2\right)\right]\)
\(A=\left[2.5+...+2^{57}.5\right]+\left[2^2.5+...+2^{58}.5\right]\)
\(A=5\left(2+2^5+...+2^{57}\right)+5\left(2^2+2^6+..+2^{58}\right)\) chia hết cho 5