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\(S=1+3+3^2+3^3+...+3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(\Rightarrow S=1.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=\left(1+...+3^{96}\right).\left(1+3+9+27\right)=\left(1+...+3^{96}\right).40\)
\(\Rightarrow S⋮40\)
S=1+3+32+33+33+...+399
=(1+3)+(32+33)+...+(398+399)
=1*(1+3)+32(1+3)+...+398(1+3)
=1*4+32*4+...+398*4
=4*(1+32+...398) chia hết 4
\(S=1+3+3^2+3^3+...+3^{99}\)
\(S=3^0+3^1+3^2+3^3+...+3^{99}\)
\(S=3^0.\left(1+3\right)+3^2.\left(1+3\right)+...+3^{98}.\left(1+3\right)\)
\(S=3^0.4+3^2.4+...+3^{98}.4\)
\(S=\left(3^0+3^2+...+3^{98}\right).4⋮4\)
\(\Rightarrowđpcm\)
a) \(\Rightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+.....+\left(3^{88}+3^{99}\right)\)
\(\Rightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+......+3^{88}\left(1+3\right)\)
\(\Rightarrow A=1.4+3^2.4+..........+3^{88}.4\)
\(\Rightarrow A=4.\left(1+3^2+.........+3^{88}\right)\)
Vậy A chia hết cho 4 ĐPCM
b) \(\Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)\)\(+......+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(\Rightarrow A=1\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+\)\(....+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=1.40+3^4.40+.......+3^{96}.40\)
\(\Rightarrow A=40.\left(1+3^4+....+3^{96}\right)\)
Vậy A chia hết cho 40 ĐPCM
\(S=4+3^2+3^3+3^4+.....+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(=\left(1+3+3^2+3^3\right).\left(1+3^4+...+3^{96}\right)\)
\(=40\left(1+3^4+...+3^{96}\right)\) \(⋮40\) (đpcm)
xét \(3S=12+3^3+3^4+....+3^{100}\)
nên 3S-S=2S=\(3^{100}-3^2-4+12=3^{100}-1\)
=>S=\(\frac{3^{100}-1}{2}\)
Ta thấy \(3^2\equiv-1\left(mod5\right)\)nên \(3^{100}\equiv1\left(mod5\right)=>S⋮5\) (1)
ta có\(3^4\equiv1\left(mod16\right)\)nên \(3^{100}\equiv1\left(mod16\right)\)=>\(S⋮8\) (2)
từ (1) (2) =>S\(⋮40\left(đpcm\right)\)
a) S=1-3+32-33+...+398-399
=>S=(1-3+32-33)+(34-35+36-37)+(38-39+310-311)+...+(396-397+398-399)
=>S=-20+34.(1-3+32-33)+38.(1-3+32-33)+...+396.(1-3+32-33)
=>S=-20+34.(-20)+38.(-20)+...+396.(-20)
=>S=-20.(1+34+38+...+396)
=>S chia hết cho -20
b) S=S = 1 - 3 + 32 - 33 + ... + 398 - 399
=>3S=3-32+33-34+...+399-3100
=>3S+S=(3-32+33-34+...+399-3100)+(1-3+32-33+...+398-399)
=>4S=1-3100
=>S=1-3100 /4
(1+3)+32(1+3+32+33)+36(1+3+32+33)+...+396(1+3+32+33)
=4+32.40+36.40+....+396.40
=4+(32+36+....+396).40:40;4+(32+36+....396).40:4
a tong S co 100 so hang, nhom thanh 25 nhom moi nhom co bon so hang, tong chia het cho -20
b) S = 1 - 3 + 32 - 33 + ... + 398 - 399
3S= 3 - 32 + 33 - ...398 + 399 - 3100
cong tung ve cua hai danh thuc ta duoc
4S= 1- 3100 ; S = 1 - 3100/ 4
S la mot so nguyen nen 1 - 3100 chia het cho 4 hay 3100 - 1 chia het cho 4 suy ra 3100 chia het cho 4 du 1
S=(1-3+32-33)+.............+(396-397+398-399)
S=(-20)+.......................+396.(1-3+32-33)
S=(-20)+...............+396.(-20)
S=(1+34+............+396).(-20) chia hết cho -20(đpcm)
b,3S=3-32+33-34+..............+399-3100
3S+S=1-3100
4S=1-3100
S=\(\frac{1-3^{100}}{4}\)
\(S=1+3+3^2+3^3+3^4+...+3^{99}\)
\(S=3^0+3^1+3^2+3^4+...+3^{99}\)
\(\Rightarrow S=3^0.\left(1+3+9+27\right)+...+3^{96}.\left(1+3+9+27\right)\)
\(\Rightarrow S=3^0.40+...+3^{96}.40\)
\(\Rightarrow S=\left(3^0+...+3^{96}\right).40⋮40\)
\(\Rightarrowđpcm\)
S=1+3+32+...+399
=(1+3+32+33)+.....+(396+397+398+399)
=1*(1+3+32+33)+....+396*(1+3+32+33)
=1*(1+3+9+27)+...+396*(1+3+9+27)
=1*40+....+396*40
=40*(1+...+396) chia hết 40
Đpcm