Bài 1:
a) Đặt A = 1 + 7 + 72 + 73 + ... + 72016
7A = 7 + 72 + 73 + 74 + ... + 72017
7A - A = (7 + 72 + 73 + 74 + ... + 72017) - (1 + 7 + 72 + 73 + ... + 72016)
6A = 72017 - 1
\(A=\frac{7^{2017}-1}{6}\)
b) Đặt B = 1 + 4 + 42 + 43 + ... + 42017
4B = 4 + 42 + 43 + 44 + ... + 42018
4B - B = (4 + 42 + 43 + 44 + ... + 42018) - (1 + 4 + 42 + 43 + ... + 42017)
3B = 42018 - 1
\(B=\frac{4^{2018}-1}{3}\)
Bài 2:
a) Ta có: \(14\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}\equiv1\left(mod13\right)\)
\(\Rightarrow14^{14}-1⋮13\left(đpcm\right)\)
b) Ta có: \(2015\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}\equiv1\left(mod2014\right)\)
\(\Rightarrow2015^{2015}-1⋮2014\left(đpcm\right)\)
Sorry mình thiếu 1+7+72+73+...+72016 câu dưới cũng thiếu 4 nha
a)A=1+7+72+73+...+72016
7A=7(1+7+72+73+...+72016)
7A=7+72+...+72017
7A-A=(7+73+...+72017)-( 1+72+73+...+72016)
6A=22017-1
\(A=\frac{2^{2017}-1}{6}\)
b)B=1+4+42+43+...+42017
4B=4(1+4+42+43+...+42017)
4B=4+42+...+42018
4B-B=(4+42+...+42018)-(1+4++...+42017)
3B=42018-1
\(B=\frac{4^{2018}-1}{3}\)