Question

[hồng nguyễn]

đặt a = 2 + 4 +4^2 +...+ 4^2022 + 4^2023 tim số dư khi chia a cho 7

Answer 1

[HT.Phong (9A5)]

\(A=2+4+4^2+...+4^{2022}+4^{2023}\)

\(A=2+2^2+\left(2^2\right)^2+\left(2^2\right)^3+...+\left(2^2\right)^{2022}+\left(2^2\right)^{2023}\)

\(A=2+2^2+2^4+2^6+...+2^{4046}\)

\(A=2+2^4+\left(2^6+2^8+2^{10}\right)+\left(2^{12}+2^{14}+2^{16}\right)+...+\left(2^{4042}+2^{4044}+2^{4046}\right)\)

\(A=2+2^4+2^6\cdot\left(1+4+16\right)+2^{12}\cdot\left(1+4+16\right)+...+2^{4042}\cdot\left(1+4+16\right)\)

\(A=2+2^4+2^6\cdot21+2^{12}\cdot21+...+2^{4042}\cdot21\)

\(A=2+16+21\cdot\left(2^6+2^{12}+...+2^{4042}\right)\)

\(A=4+14+21\cdot\left(2^6+2^{12}+...+2^{4042}\right)\)

\(A=4+7\cdot\left[2+3\cdot\left(2^6+2^{12}+...+2^{4042}\right)\right]\)

Mà: \(7\cdot\left[2+3\cdot\left(2^6+2^{12}+...+2^{4042}\right)\right]\) ⋮ 7

⇒ \(A=4+7\cdot\left[2+3\cdot\left(2^6+2^{12}+...+2^{4042}\right)\right]\) : 7 dư 4 

Vậy: ... 

Answer 2

[Kiều Vũ Linh]

A = 2 + 4 + 4² + ... + 4²⁰²² + 4²⁰²³

= 2 + 4 + (4² + 4³ + 4⁴) + (4⁵ + 4⁶ + 4⁷) + ... + (4²⁰²¹ + 4²⁰²² + 4²⁰²³)

= 6 + 4.(4 + 4² + 4³) + 4⁴.(4 + 4² + 4³) + ... + 4²⁰²⁰.(4 + 4² + 4³)

= 6 + 4.84 + 4⁴.84 + ... + 4²⁰²⁰.84

= 6 + 84.(4 + 4⁴ + ... + 4²⁰²⁰)

= 6 + 7.12.(4 + 4⁴ + ... + 4²⁰²⁰)

Mà 7.12.(4 + 4⁴ + ... + 4²⁰²⁰)

⇒ 6 + 7.12.(4 + 4⁴ + ... + 4²⁰²⁰) chia 7 dư 6

Vậy A chia 7 dư 6