Question

[Minamoto Shizuka]

Cho A=1+2+2^2+2^3+...+2^2013+2^2014         a,Tính A           b,CTR:A chia hết cho 31

Answer 1

[Nguyễn Phạm Hồng Anh]

a,A = 1 + 2 + 2² + 2³ +.... + 2²⁰¹³ + 2²⁰¹⁴

2A = 2 + 2² + 2³ + ...... + 2²⁰¹³ + 2²⁰¹⁴ + 2²⁰¹⁵

A  = ( 2 + 2² + 2³ + ..... + 2²⁰¹³ + 2²⁰¹⁴ + 2²⁰¹⁵ ) - ( 1 + 2 + 2² + 2³ + ..... + 2²⁰¹³ + 2²⁰¹⁴ )

A = 2²⁰¹⁵ - 1

b, A = 1 + 2 + 2² + 2³ + ... + 2²⁰¹³ + 2²⁰¹⁴

       = ( 1 + 2 + 2² + 2³ + 2⁴ ) + .... + ( 2²⁰¹⁰ + 2²⁰¹¹ + 2²⁰¹² + 2²⁰¹³ + 2²⁰¹⁴ )

       = 31 + ..... + 2²⁰¹⁰.( 1 + 2 + 2² + 2³ + 2⁴ )

       = 31 + ..... + 2²⁰¹⁰ . 31

       = 31.1 + ..... + 2²⁰¹⁰ . 31

       = 31. ( 1 + .... + 2²⁰¹⁰ ) chia hết cho 31

=> A chia hết cho 31

Answer 2

[Không Tên]

a)   \(A=1+2+2^2+2^3+....+2^{2014}\)

\(\Leftrightarrow\)\(2A=2+2^2+2^3+2^4+...+2^{2015}\)

\(\Leftrightarrow\)\(2A-A=\left(2+2^2+2^3+...+2^{2015}\right)-\left(1+2+2^2+...+2^{2014}\right)\)

\(\Leftrightarrow\)\(A=2^{2015}-1\)

b)    \(A=1+2+2^2+2^3+...+2^{2014}\)

\(=\left(1+2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8+2^9\right)\)\(+...+\left(2^{2010}+2^{2011}+2^{2012}+2^{2013}+2^{2014}\right)\)

\(=\left(1+2+2^2+2^3+2^4\right)+2^5\left(1+2+2^2+2^3+2^4\right)\)\(+...+2^{2010}\left(1+2+2^2+2^3+2^4\right)\)

\(=\left(1+2+2^2+2^3+2^4\right)\left(1+2^5+...+2^{2010}\right)\)

\(=31\left(1+2^5+...+2^{2010}\right)\)  \(⋮31\)