Question

cho S=3+3²3³+...+3¹⁰⁰

^{chứng minh rằng S chia hết cho 40}

Answer 1

\(S=3+3^2+3^3+...+3^{100}\)

\(S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(S=40.3+...+3^{96}\left(3+3^2+3^3+3^4\right)\)

\(S=40.3+...+3^{96}.40.3\)

\(S=40.3.\left(3^4+...+3^{96}\right)\)chia hết 40

Answer 2

Ta có: S = 3 + 3² + 3³ + ...... + 3¹⁰⁰

=> 3S = 3² + 3³ + 3³ +...... + 3¹⁰¹

=> 3S - S = 3¹⁰¹ - 3

=> 2S = 3¹⁰¹ - 3

=> S = \(\frac{3^{101}-3}{2}\)

Answer 3

Cho mk xin lỗi k nhầm đề:

 Ta có: S = 3 + 3² + 3³ + ..... + 3¹⁰⁰

=> S = (3 + 3² + 3³ + 3⁴) + ..... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰)

=> S = 3.(1 + 3 + 9 + 27) + ..... + 3⁹⁷.(1 + 3 + 9 + 27)

=> S = 3.40 + ..... + 3⁹⁷.40

=> S = 40.(3 + .... + 3⁹⁷) chia hết cho 40 (đpcm)