Question

[H. Nhiên Phạm]

1.Chứng tỏ:

A=2+2²+2³+2⁴+...+2²⁰⁰⁴ chia hết cho 3, cho 7 và cho 15.

B=1+3+3²+3³+...+3⁹⁹chia hết cho 40.

2.Chứng tỏ chia có dư:

L=1+2+2²+2³⁺...+2²⁰⁰⁹+2²⁰¹⁰ chia 7 dư 1.

*P/S:Giải từng bước nha!

Answer 1

[Nguyễn Huy Tú]

Bài 1:

a) +) \(A=2+2^2+...+2^{2004}\)

\(\Rightarrow A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2003}+2^{2004}\right)\)

\(\Rightarrow A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2003}\left(1+2\right)\)

\(\Rightarrow A=2.3+2^3.3+...+2^{2003}.3\)

\(\Rightarrow A=\left(2+2^3+...+2^{2003}\right).3⋮3\)

\(\Rightarrow A⋮3\left(đpcm\right)\)

+) \(A=2+2^2+...+2^{2004}\)

\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{2002}+2^{2003}+2^{2004}\right)\)

\(\Rightarrow A=2\left(1+2+2^2\right)+...+2^{2002}\left(1+2+2^2\right)\)

\(\Rightarrow A=2.7+...+2^{2002}.7\)

\(\Rightarrow A=\left(2+...+2^{2002}\right).7⋮7\)

\(\Rightarrow A⋮7\left(đpcm\right)\)

+) \(A=2+2^2+....+2^{2004}\)

\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)

\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{2001}\left(1+2+2^2+2^3\right)\)

\(\Rightarrow A=2.15+...+2^{2001}.15\)

\(\Rightarrow A=\left(2+...+2^{2001}\right).15⋮15\)

\(\Rightarrow A⋮15\left(đpcm\right)\)

b) \(B=1+3+3^2+...+3^{99}\)

\(\Rightarrow B=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)

\(\Rightarrow B=\left(1+3+9+27\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow B=40+...+3^{96}.40\)

\(\Rightarrow B=\left(1+...+3^{96}\right).40⋮40\)

\(\Rightarrow B⋮40\left(đpcm\right)\)

Answer 2

[Đinh Hải Ngọc]

đpcm là điều phải chứng minh !