\(A=7+7^2+7^3+........+7^{2016}\)

\(A=7\left(1+7+7^2+7^3+........+7^{2012}+7^{2013}+7^{2014}+7^{2015}\right)\)

\(A=7\left[\left(1+7+7^2+7^3\right)+........+\left(7^{2012}+7^{2013}+7^{2014}+7^{2015}\right)\right]\)

\(A=7\left[\left(1+7+7^2+7^3\right)+........+7^{2012}\left(1+7+7^2+7^3\right)\right]\)

\(A=7\left[400+........+7^{2012}.400\right]\)

\(A=7.400\left(1+7^4+7^8+7^{12}+......+7^{2012}\right)⋮400\)

Vì \(20^2=400\) nên \(A⋮20^2\left(dpcm\right)\)

chị kết bạn với em nha gửi lời kết bn với em nhé

j zậy em hả 

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+...+7^{2016 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

\(B=\frac{4^{2018}-1}{3}\)

ngu

7S=7+7^2+7^3+7^4+...+7^2016

=>7S-S=(7+7^2+7^3+7^4+...+7^2016)-(1+7+7^2+7^3+...+7^2015)

=>6S=7^2016-1

=>6S+1=7^2016-1+1=7^2016(đpcm)