Chung to rang 1+5+52+........+5403+5404 Chia hết cho 31
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Đặt \(A=1+5+5^2+5^3+...+5^{402}+5^{403}+5^{404}\)
\(\Rightarrow A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{399}+5^{400}+5^{401}\right)+\left(5^{402}+5^{403}+5^{404}\right)\)
\(\Rightarrow A=31.1+31.5^3+...+31.5^{402}\)
\(\Rightarrow A=31\left(1+5^3+5^6+...+5^{402}\right)\)
\(\Rightarrow A⋮31\left(đpcm\right)\)
\(\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\\ =31+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\\ =31+5^3.31+...+5^{402}.31\\ =31.\left(1+5^3+...+5^{402}\right)⋮31\left(DPCM\right)\)
1+5+52+....+5404
= (1+5+52) + (53+54+55) + .......+ (5402 + 5403 + 5404)
= 1(1+5+52) + 53(1+5+52) +......+ 5402(1+5+52)
= 1. 31 + 53. 31 +......+5402. 31
= 31(1 + 53 + ......... + 5402) chia hết cho 31 (đpcm)
1+5+52+....+5404
= (1+5+52) + (53+54+55) + .......+ (5402 + 5403 + 5404)
= 1(1+5+52) + 53(1+5+52) +......+ 5402(1+5+52)
= 1. 31 + 53. 31 +......+5402. 31
= 31(1 + 53 + ......... + 5402) chia hết cho 31 (đpcm)
Ta có\(5^{2012}+5^{2011}+5^{2010}=5^{2010}\left(25+5+1\right)=5^{2010}\cdot31⋮31\)(đpcm)
Ta có :
A=2+2^2+2^3+2^4+...+2^{99}+2^{100}A=2+22+23+24+...+299+2100
=> A=\left(2+2^2+2^3+2^4+2^5\right)+....\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)A=(2+22+23+24+25)+....(296+297+298+299+2100)
=> A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)A=2(1+2+22+23+24)+...+296(1+2+22+23+24)
=> A=2.31+...+2^{96}.31A=2.31+...+296.31
=> A=\left(2+...+2^{96}\right)31A=(2+...+296)31chia hết cho 31