Question

Cho B = 3 + 3 mũ 2 +3 mũ 3 +...+3 mũ 2020

CMR:B chia hết cho 4

 

Answer 1

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Answer 2

Ta có: 

\(B=3+3^2+3^3+...+3^{2020}\)

\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2019}+3^{2020}\right)\)

\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)

\(B=3\cdot4+3^3\cdot4+...+3^{2019}\cdot4\)

\(B=4\cdot\left(3+3^3+...+3^{2019}\right)\) chia hết cho 4

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Answer 3

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Answer 4

\(B=3+3^2+3^3+...+3^{2020}\)

\(3B=3.(3+3^2+3^3+...+3^{2020}\)\()\)

\(3B=3^2+3^3+3^4+...+3^{2021}\)

\(3B-B=\left(3^2+3^3+3^4+...+3^{2021}\right)-\left(3+3^2+3^3+...+3^{2020}\right)\)

\(2B=3^{2021}-3\)

\(B=\frac{3^{2021}-3}{2}\)

Vậy: \(B=\frac{3^{2021}-3}{2}\)

Answer 5

B = 3 + 3² + 3³ + ....+3²⁰²⁰

B = ( 3 + 3² ) + ( 3³ + 3⁴ ) + ..... + ( 3²⁰¹⁹ + 3²⁰²⁰ )

B = 3 . ( 1 + 3 ) + 3³ . 4  + .... + 3²⁰¹⁹ . 4

B = 4 . ( 3 + 3³ + .... + 3²⁰¹⁹ ) chia hết cho 4

=> Đpcm