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![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2E=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow4E< 3\)
\(\Rightarrow E< \frac{3}{4}\left(đpcm\right)\)
Bài 1:
Ta có: \(3+3^2+3^3+...+3^{100}\)
\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=120+3^5\left(3+3^2+3^3+3^4\right)+....+3^{96}\left(3+3^2+3^3+3^4\right)\)
\(=120+3^5.120+...+3^{96}.120\)
\(=120.\left(1+3^5+.....+3^{96}\right)\)
\(\Rightarrow3+3^2+3^3+3^4+....+3^{100}\)chia hết cho 120 (vì có chứa thừa số 120)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
a)\(B=3+3^3+3^5+3^7+.....+3^{1991}\)
\(\Leftrightarrow B=3\left(1+3^2+3^4+3^6+.....+3^{1990}\right)\)
Vì \(3\left(1+3^2+3^4+3^6+.....+3^{1990}\right)\)chia hết cho 3 nên \(B⋮3\)
\(B=3+3^3+3^5+3^7+.....+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+.....+\left(3^{1988}+3^{1989}+3^{1990}+3^{1991}\right)\)
\(\Leftrightarrow B=3\left(1+3^2+3^4+3^6\right)+.....+3^{1988}\left(1+3^2+3^4+3^6\right)\)
\(\Leftrightarrow B=3.820+.....+3^{1988}.820\)
\(\Leftrightarrow B=3.20.41+.....+3^{1988}.20.41\)
Vì \(3.20.41+.....+3^{1988}.20.41\) chia hết cho 41 nên \(B⋮41\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Giải :
Ta có: \(E=5+5^2+5^3+5^4+...+5^{97}+5^{98}+5^{99}+5^{100}\) \(\Leftrightarrow E=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{97}+5^{98}\right)+\left(5^{99}+5^{100}\right)\)
\(\Leftrightarrow E=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^{97}.\left(1+5\right)+5^{99}.\left(1+5\right)\)
\(\Leftrightarrow E=5.6+5^3.6+...+5^{97}.6+5^{99}.6\)
\(\Leftrightarrow E=6.\left(5+5^3+...+5^{97}+5^{99}\right)\)
\(\Rightarrow E⋮6\)
Do \(E⋮6\)nên \(E\div6\)dư 0
Vậy \(E\div6\)có số dư bằng \(0\)
Bài 2:
Giải :
Ta có: \(n.\left(n+2\right).\left(n+7\right)\)
\(=\left(n^2+2n\right).\left(n+7\right)\)
\(=n^3+2n^2+7n^2+14n\)
\(=n^3+9n^2+14n\)
\(=n.\left(n^2+9n+14\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+4+4^2+...+4^59A=1+4+4^2+...+4^59
A=(1+4)+(4^2+4^3)+...+(4^58+4^59)A=(1+4)+(4^2+4^3)+...+(4^58+4^59)
A=(1+4)+4^2(1+4)+...+4^58(1+4)A=(1+4)+4^2(1+4)+...+4^58(1+4)
A=5+4^2.5+...+4^58.5A=5+4^2.5+...+4^58.5
A=5(1+4^2+...+4^48)A=5(1+4^2+...+4^58)
A=5(1+4^2+...+4^58) chia hết cho 5
vậy A chia hết cho 5
A=1+4+4^2+...+4^59A=1+4+4^2+...+4^59
A=(1+4+4^2)+(4^3+4^4+4^5)+...+(4^57+4^58+4^59)A=(1+4+4^2)+(4^3+4^4+4^5)+...+(4^57+4^58+4^59)
A=(1+4+4^2)+4^3(1+4+4^2)+...+4^57(1+4+4^2)A=(1+4+4^2)+4^3(1+4+4^2)+...+4^57(1+4+4^2)
A=21+4^3.21+...+4^57.21A=21+4^3.21+...+4^57.21
A=21(1+4^3+...+4^57)A=21(1+4^3+...+4^57)
A=21(1+4^3+...+4^57) chia hết cho 21
vậy A chia hết cho 21
mik làm xong rồi nhớ k cho mik nha mik cảm ơn
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1+3+\(3^2\)+\(3^3\)+....+\(3^{1991}\)
B=1+3+\(3^2\)+\(3^3\)+....+\(3^{1991}\)
=(1+3+\(3^2\)+\(3^3\))+(\(3^4\)+\(3^5\)+\(3^6\)+\(3^7\))+.....+(\(3^{1988}\)+\(3^{1989}\)+\(3^{1990}\)+\(3^{1991}\))
=(1+\(3^4\))(1+3+\(3^2\)+\(3^3\))(\(3^8\)+....+\(3^{1988}\))
=82.(1+3+\(3^2\)+\(3^3\))(\(3^8\)+....+\(3^{1988}\))
Vì 82⋮41
→E⋮41
→B⋮41(đpcm)
Bạn tham khảo nha:
B=1+3+32+33+....+31991B=1+3+32+33+....+31991
=(1+3+32+33)+(34+35+36+37)+.....+(31988+31989+31990+31991)=(1+3+32+33)+(34+35+36+37)+.....+(31988+31989+31990+31991)
=(1+3+32+33)+34(1+3+32+33)+....+31988(1+3+32+33)=(1+3+32+33)+34(1+3+32+33)+....+31988(1+3+32+33)
=(1+3+32+33)+(1+34+....+31988)=(1+3+32+33)+(1+34+....+31988)
=(1+34)(1+3+32+33)(38+....+31988)=(1+34)(1+3+32+33)(38+....+31988)
=82.(1+3+32+33)(38+....+31988)=82.(1+3+32+33)(38+....+31988)
Vì 82⋮4182⋮41
→82.(1+3+32+33)(38+....+31988)⋮41→82.(1+3+32+33)(38+....+31988)⋮41
→B⋮41(đpcm)
Ta có:
3! chia hết cho 3.
4! chia hết cho 3.
........
100! chia hết cho 3.
Vì mỗi số trên chia hết cho 3 nên tổng chia hết cho 3.
Lại có:
1!=1
2!=1.2=2
1!+2!=3 chia hết cho 3.
Vậy E .......................
Chúc em học tốt^^
thanks nha chi