Question

[Minamoto Shizuka]

Chứng minh rằng

7¹ + 7² + 7³+ 7⁴ + ......... +7^4n-1 +7⁴ⁿ chia hết cho 400

Answer 1

[Lightning Farron]

\(7^1+7^2+...+7^{4n-1}+7^{4n}\)

\(=\left(7^1+7^2+7^3+7^4\right)+...+\left(7^{4n-3}+7^{4n-2}+7^{4n-1}+7^{4n}\right)\)

\(=7^1\left(1+7+7^2+7^3\right)+...+7^{4n-3}\left(1+7+7^2+7^3\right)\)

\(=7^1\cdot400+...+7^{4n-3}\cdot400\)

\(=400\left(7^1+...+7^{4n-3}\right)⋮400\)

Answer 2

[Chibi dễ thương ^ _ ^]

7¹ + 7² + 7³ + 7⁴ + ... + 7^{4n - 1} + 7⁴ⁿ

= (7¹ + 7² + 7³ + 7⁴) + (7⁵ + 7⁶ + 7⁷ + 7⁸) + ... + (7^{4n - 3} + 7^{4n - 2} + 7^{4n - 1} + 7⁴ⁿ)

= 7¹ . (1 + 7 + 7² + 7³) + 7⁵ . (1 + 7 + 7² + 7³) + ... + 7^{4n - 3} . (1 + 7 + 7² + 7³)

= 7¹ . 400 + 7⁵ . 400 + ... + 7^{4n - 3} . 400

= 400 . (7¹ + 7⁵ + ... + 7^{4n - 3})

Vì 400 \(⋮\)400 nên suy ra 400 . (7¹ + 7⁵ + ... + 7^{4n - 3}) \(⋮\)400

Vậy ....

~.~