Question

Chứng tỏ  1 + 7+ 7² + 7³ + ... + 7¹⁰¹ chia hết cho 8

Answer 1

\(1+7+7^2+...+7^{101}\)

Nhóm các cặp số lại với nhau :

\(\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{100}+7^{101}\right)=8+7^2\left(1+7\right)+7^4\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(\Leftrightarrow8\cdot\left(1+7^2+7^4+...+7^{100}\right)⋮8\)

Answer 2

D=(1+7)+7²=(1+7)+......+7¹⁰⁰(1+7)

D=8+7².8+.........+7¹⁰⁰.8

D=8(1+7²+...+7¹⁰⁰) chia hết cho 8

Vậy D chia hết cho 8

Answer 3

\(1+7+7^2+7^3+...+7^{101}\)

\(=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{100}-7^{101}\right)\)

\(=8+7^2\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(=8+7^2\cdot8+...+7^{100}\cdot8\)

\(=8\left(1+7^2+...+7^{100}\right)⋮8\)

\(\Rightarrow1+7+7^2+7^3+....+7^{101}⋮8\left(đpcm\right)\)