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đặt biểu thức ban đầu là A, 42020+42019+...+4+1=B
4B=42021 +42020 +42019+...+42+4
3B=4B-B=42021-1 => B= (42021-1)/3
A=75B+25=75(42021-1)/3 + 25= 25(42021-1)+25=25(42021-1+1)=25.42021=100.42020
=> A chia hết cho cả 100 và 42021
mặt khác A=25.42021=42021.(24+1)=24.42021+42021=6.42022+42021
vì 42021<42022 nên A chia 42022 dư 42021
tick cho mk nha!!!!!!!!
M=75.(42013+42012+…..+43+42+1)+25
=75.42013+75.42012+……+75.43+75.42+75.1+25
=75.42013+75.42012+……+75.43+75.42+75+25
=75.42013+75.42012+……+75.43+75.42+100
=3.(25.4).42012+3.(25.4).42011+…..+3.(25.4).42+3.(25.4).4+100
=3.100.42012+3.100.42011+…..+3.100.42+3.100.4+100
=100.(3.42012+3.42011+…..+3.42+3.4+1)
Vì 100 chia het 100 nen 100.(3.42012+3.42011+…..+3.42+3.4+1) chia het 100
Vậy M chia het 100
Đặt A = 42016 + 42015 + ... + 42 + 4 + 1
=> A = 4.k + 1 (k \(\in\)N*)
P = 75.(4.k + 1) + 25
P = 75.4k + 75 + 25
P = 300.k + 100
P = 100.(3.k + 1) chia hết cho 100 (đpcm)
\(A=75\left[4\left(4^{2006}+4^{2005}+...+4+1\right)+1\right]+25\)
\(A=300\left(4^{2006}+4^{2005}+...+4+1\right)+75+25\)
\(A=300\left(4^{2006}+4^{2005}+...+4+1\right)+100\)
\(A=100\left[3\left(4^{2006}+4^{2005}+...+4+1\right)+1\right]⋮100\)
\(3,1+5^2+5^4+...+5^{26}\)
\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)
\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)
\(=26+5^4.26+...+5^{24}.26\)
\(=26\left(5^4+...+5^{24}\right)\)
Vì \(26⋮26\)
\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)
\(4,1+2^2+2^4+...+2^{100}\)
\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)
\(=21+2^6.21...+2^{98}.21\)
\(=21\left(2^6+...+2^{98}\right)\)
Có : \(21\left(2^6+...+2^{98}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)
B=25.3.(42003+42002+22001+.......+42+4+1)+25
B=25.[4.(42003+42002+22001+.......+42+4+1)-(42003+42002+22001+.......+42+4+1)]+25
B=25.[(42004+42003+42002+22001+.......+42+4)-(42003+42002+22001+.......+42+4+1)]+25
B=25.(42004-1)+25
B=25.(42004-1+1)
B=25.42004
B=25.4.42003
B=100.42003
\(\Rightarrow\)B chia hết cho 100
A=75(4^2004+4^2003+...+4^24+1)+25= 75(4^2004+4^2003+...+4^24)+75+25=
=75(4^2004+4^2003+...+4^24)+100= 75*4(4^2003+4^2002...+4^23)+100=
= 300(4^2003+4^2002...+4^23)+100= 100[3(4^2003+4^2002...+4^23)+1] chia het cho 100.
Ta có M ⋮ 25 vì 75 ⋮ 25
Lại có M = 75 ( 42021 + 42020 + ... + 42 + 4 + 1 )
= 75 . 4 ( 22020 + 22019 + ... + 4 + 1 + 0,25 ) ⋮ 4 vì 4 ⋮ 4
Mà ( 25; 4 ) = 1 ⇒ M ⋮ 100
Vậy M ⋮ 100