cho A=2+2^1+2^2+2^3+2^4+...+2^100 chứng minh chia hết cho 3
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KV
8 tháng 11 2023
a) Đặt A = \(6^5.5-3^5\)
\(=\left(2.3\right)^5.5-3^5\)
\(=2^5.3^5.5-3^5\)
\(=3^5.\left(2^5.5-1\right)\)
\(=3^5.\left(32.5-1\right)\)
\(=3^5.159\)
\(=3^5.3.53⋮53\)
Vậy \(A⋮53\)
b) Đặt \(B=2+2^2+2^3+...+2^{120}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{119}.3\)
\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)
Vậy \(B⋮3\)
\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{118}.7\)
\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)
Vậy \(B⋮7\)
\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)
\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+2^6.31+...+2^{116}.31\)
\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)
Vậy \(B⋮31\)
\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)
\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)
\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)
\(=2.255+2^9.255+...+2^{113}.255\)
\(=255.\left(2+2^9+...+2^{113}\right)\)
\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)
Vậy \(B⋮17\)
KV
8 tháng 11 2023
c) Đặt C = \(3^{4n+1}+2^{4n+1}\)
Ta có:
\(3^{4n+1}=\left(3^4\right)^n.3\)
\(2^{4n}=\left(2^4\right)^n.2\)
\(3^4\equiv1\left(mod10\right)\)
\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)
\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)
\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)
\(2^4\equiv6\left(mod10\right)\)
\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)
\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)
\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)
\(\Rightarrow\) Chữ số tận cùng của C là 5
\(\Rightarrow C⋮5\)
18 tháng 11 2018
\(A=\left(2+2^2\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=2\cdot\left(1+2\right)+...+2^{99}\cdot\left(1+2\right)\)
\(A=2\cdot3+...+2^{99}\cdot3\)
\(A=3\cdot\left(2+...+2^{99}\right)⋮3\left(đpcm\right)\)
2 ý kia tương tự
18 tháng 11 2018
Giải:
Đặt S=(2+2^2+2^3+...+2^100)
=2.(1+2+2^2+2^3+2^4)+2^6.(1+2+2^2+2^3+2^4)+...+(1+2+2^2+2^3+2^4).296
=2.31+26.31+...+296.31
=31.(2+26+...+296)\(⋮\)31
ND
2
MH
3 tháng 1 2022
\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{99}\right)\) ⋮ 3 \(\left(đpcm\right)\)
3 tháng 1 2022
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{99}\right)=3\left(2+2^3+...+2^{99}\right)⋮3\)
1T
1
5 tháng 1 2022
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)=3\left(2+2^3+...+2^{99}\right)⋮3\)
TT
2
10 tháng 4 2022
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=3\left(2+...+2^{99}\right)⋮3\)