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B = (1 + 3) + (32+33)+.....+(389+390)
= 4 + 32 .(1 + 3) + .....+390.(1+3)
= 1 .4 + 32.4 + ..... +390.4
= 4.(1 + 32 + .... +390) chia hết cho 4
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
Ta có ;
S = 3 + 3 2 + 3 3 + ........ + 3 99 + 3 100
= ( 3 + 3 2 + 3 3 + 3 4 + 3 5) + .... + ( 3 96 + 3 97 + 3 98 + 3 99 + 3 100 )
= 3 ( 1 + 3 + 3 2 + 3 3 + 3 4 ) + .... + 3 96 . ( 1 + 3 + 3 2 + 3 3 + 3 4 )
= 3 . 121 + .... + 3 96 . 121
= 121 . ( 3 + .... + 3 96 ) chia hết cho 121 ( Do 121 chia hết cho 121 )
Vậy S = 3 + 3 2 + 3 3 + ........ + 3 99 + 3 100 chia hết cho 121
Mk ngĩ ra rồi
S=(1+32)+(34+36)+...+(396+398)
S=10+34.(1+32)+...+396.(1+32)
S=10+34.10+...+396.10
S=10(1+34+...+396)
có thừa số 10 chia hết cho 10 nên tích chia hết cho 10
s = 3 ^0 + 3 ^ 2 + 3^ 4+ 3 ^6 +... + 3 ^2002
9S = 3 ^4 + 3^6 + 3 ^ 2004
9S - S= 3 ^ 2004 - 1
8S = 3^2004 - 1
S = 3 ^ 2004 - 1/8
k mk nha
\(S=4+3^2+3^3+3^4+.....+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(=\left(1+3+3^2+3^3\right).\left(1+3^4+...+3^{96}\right)\)
\(=40\left(1+3^4+...+3^{96}\right)\) \(⋮40\) (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\(⋮40\left(đpcm\right)\)
\(S=3+3^2+3^3+...+3^{2019}\)
\(3S=3^2+3^3+...+3^{2019}+3^{2020}\)
\(\Rightarrow3S-S=-3+3^{2020}\)
\(\Rightarrow2S=3^{2020}-3\Rightarrow S=\frac{3^{2020}-3}{2}\)
Ta có: \(S=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2017}+3^{2018}+3^{2019}\right)\)
\(=3\left(1+3+9\right)+3^4\left(1+3+9\right)+...+3^{2017}\left(1+3+9\right)\)
\(=3.13+3^4.13+...+3^{2017}.13\)
\(=13.\left(3+3^4+...+3^{2017}\right)⋮13\)