Chứng minh rằng:
a, (1+2+2^2+2^3+...+2^7) chia hết cho 3
b, (1+2+2^2+2^3+...+2^11) chia het cho 9
c, A=2+2^2+2^3+...+2^60 chia hết cho 3;7;15
d, B=3+3^3+3^5+...+3^1991 chia hết cho 13;41
e, 10^28+8 chia hết cho 72
f, 8^8+2^20 chia hết cho 17
g, S1=5+5^2+5^3+...+5^100 chia hết cho 6
S2=2+2^2+2^3+...+2^100 chia hết cho 31
S3=16^5+2^15 chia hết cho 33
a) \(\left(1+2+2^2+...+2^7\right)\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^6+2^7\right)\)
\(=\left(1+2\right)+2^2.\left(1+2\right)+...+2^6.\left(1+2\right)\)
\(=3+2^2.3+...+2^6.3\)
\(=3.\left(1+2^2+...+2^6\right)⋮3\left(đpcm\right)\)
a) Đặt A = 1 + 2 + 22 + 23 + ... + 27
Ta có:
A = 1 + 2 + 22 + 23 + ... + 27
\(\Rightarrow\)2A = 2 + 22 + 23 + 24 + ... + 28
\(\Rightarrow\)A = 28 - 1 = 255
Vì 255\(⋮\)3\(\Rightarrow\)2 + 22 + 23 + 24 + ... + 28\(⋮\)3
\(\Rightarrow\)ĐPCM
