Question

Cho A = 7 + 7³ + 7⁵ + ... + 7¹⁹⁹⁹ Chứng minh A chia hết cho 35.

Answer 1

Ta có:\(A=7+7^3+7^5+7^7+...+7^{1998}+7^{1999}\)

\(=\left(7+7^3\right)+\left(7^3+7^5\right)+...+\left(7^{1998}+7^{1999}\right)\)

\(=\left(7+7^3\right)+7^2.\left(7+7^3\right)+...+7^{^{1997}}.\left(7+7^3\right)\)

\(=350+7^2.350+...+7^{1997}.350\)

\(=350.\left(1+7^2+...+7^{1997}\right)\)

\(=35.10.\left(1+7^2+...+7^{1997}\right)\)

VÌ 35.10.(1+7²+...+7¹⁹⁹⁷) CHIA HẾT CHO 35 

NÊN A CHIA HẾT CHO 35

Answer 2

A=7 + 7³ + 7⁵ +... + 7¹⁹⁹⁹=(7 + 7²) + (7^{5 +} 7⁷)+...+(7¹⁹⁹⁷ + 7¹⁹⁹⁹)

A=7(1 + 7²) + 7⁵⁽1 + 7²)+...+7¹⁹⁹⁷(1 + 7²)

A=7 x 50 + 7⁵ +...+ 7 =7 x 7¹⁹⁹⁷ x 50

=>A chia hết cho 5 (1)

A=7 + 7³ + 7⁵ +....+ 7¹⁹⁹⁹=7 x(7⁰ + 7² + 7⁴  + ...7¹⁹⁹⁸)

=>A Chia hết cho 7(2)

Mà ƯCLN(5,7)=1=>A Chia hết cho 35

Answer 3

A chia hết 35 nhé cj