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Đặt \(A_1=\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)
Ta có: \(A_1=\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)
\(\Leftrightarrow4A_1=4+4^2+4^3+...+4^{2017}+4^{2018}\)
Lấy \(4A_1-A_1\)ta có:
\(4A_1-A_1=\left(4+4^2+4^3+...+4^{2017}+4^{2018}\right)-\left(1+4+4^2+...+4^{2016}+4^{2017}\right)\)
\(\Leftrightarrow3A_1=4^{2018}-1\)
\(\Leftrightarrow A_1=\frac{4^{2018}-1}{3}\)
Thay \(A_1=\frac{4^{2018}-1}{3}\)vào biểu thức A, ta có:
\(A=75.\left(\frac{4^{2018}-1}{3}\right)+25\)
\(\Leftrightarrow A=25.\left(4^{2018}-1\right)+25\)
\(\Leftrightarrow A=25.4^{2018}⋮4^{2018}\)
Vậy \(A⋮4^{2018}\)
chúc bn hok tốt
đặt \(S=1+4+4^2+......+4^{1999}\)
\(\Rightarrow4S=4+4^2+4^3+....+4^{2000}\)
\(\Rightarrow4S-S=\left(4+4^2+4^3+....+4^{2000}\right)-\left(1+4+4^2+.....+4^{1999}\right)\)
\(\Rightarrow3S=4^{2000}-1\Rightarrow S=\frac{4^{2000}-1}{3}\)
Khi đó \(A=75.S=75.\frac{4^{2000}-1}{3}=\frac{75.\left(4^{2000}-1\right)}{3}=\frac{75}{3}.\left(4^{2000}-1\right)=25.\left(4^{2000}-1\right)=25.4^{2000}-25\)
Ta có: 42000-1=(44)500-1=(...6)-1=....5
=>25.42000-25=25.(....5)-25=(...5)-25=....0 chia hết cho 100
Vậy ta có điều phải chứng minh
75 chia hết cho 25.
42007 + ... + 4 + 1 chia 4 dư 1 hay không chia hết cho 4
=> 75(42007 + ... + 4 + 1) không chia hết cho 100.
\(A=1+3^2+3^4+...+3^{100}\)
\(9A=3^2+3^4+3^6+...+3^{102}\)
\(8A=3^{102}-1\)
\(\Rightarrow8A-26=3^{102}-1-26=3^{102}-27\)
Vì \(3^{102}-27⋮3\)(1)
\(3^{102}-27⋮2\)(\(3^{102}-27\)là số chẵn ) (2)
\(3^{102}-27=9\left(3^{100}-3\right)\)\(\Rightarrow3^{102}-27⋮9\)(3)
Từ (1) , (2), (3) \(\Rightarrow8A-26⋮54\)\(\left(\left(2,3,9\right)=1\right)\)
vậy ...
\(A=1+3^2+3^4+...+3^{100}\)
\(\Leftrightarrow3^2A=3^2\left(1+3^2+3^4+....+3^{100}\right)\)
\(\Leftrightarrow9A=3^2+3^4+3^6+...+3^{102}\)
\(\Leftrightarrow9A-A=\left(3^2+3^4+3^6+....+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(\Leftrightarrow8A=3^{102}-1\)
\(\Leftrightarrow8A-26=3^{102}-1-26=3^{102}-27\)
Ta có: \(3^{102}⋮3;27⋮3\Rightarrow3^{102}-27⋮3\left(1\right)\)
\(3^{102}-27⋮2\left(2\right)\)(3^102 -27 là số lẻ)
\(3^{102}-27=\left(3^2\right)^{51}-27=9^{51}-27⋮9\left(3\right)\)
(1)(2)(3) => 8A-26 chia hết cho 54 (đpcm)
Đặt \(B=1+4+4^2+...+4^{1998}+4^{1999}\)
\(\Rightarrow4B=4+4^2+4^3+...+4^{1999}+4^{2000}\)
\(\Rightarrow4B-B=\left(4+4^2+4^3+...+4^{2000}\right)-\left(1+4+4^2+...+4^{1999}\right)\)
\(\Rightarrow3B=4^{2000}-1\)
\(\Rightarrow B=\dfrac{4^{2000}-1}{3}\)
Khi đó ta có:
\(A=75.B=75.\dfrac{4^{2000}-1}{3}=\dfrac{75.\left(4^{2000}-1\right)}{3}=\dfrac{75}{3}.\left(4^{2000}-1\right)=25.\left(4^{2000}-1\right)=25.4^{2000}-25\)
Ta có: \(4^{2000}-1=\left(4^4\right)^{500}-1=\left(...6\right)-1=...5\)
\(\Rightarrow25.4^{2000}-25=25.\left(...5\right)-25=\left(...5\right)-25=...0⋮100\left(đpcm\right)\)
Ta có:
\(A=75.\left(4^{1999}+4^{1998}+...+4^2+4+1\right)+25\)
\(A=25.3.\left(4^{1999}+4^{1998}+...+4^2+4+1\right)+25\) \(A=25.\left(4-1\right).\left(4^{1999}+4^{1998}+...+4^2+4+1\right)+25\)
\(A=25.\left(4^{2000}+4^{1999}+...+4^3+4^2+4-4^{1999}-4^{1998}-...-4^2-4-1\right)+25\)\(A=25.\left(4^{2000}-1\right)+25\)
\(A=25.\left(4^{2000}-1+1\right)\)
\(A=25.4^{2000}=25.4.4^{1999}=100.4^{1999}\)Vây:A là số chia hết cho 100
