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a: \(G=8^8+2^{20}\)
\(=2^{24}+2^{20}\)
\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)
b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)⋮15\)
c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
\(E=1+3+3^2+3^3+...+3^{1991}\)
\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)
\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)
\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2E=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4E=3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow4E< 3\)
\(\Rightarrow E< \frac{3}{4}\left(đpcm\right)\)
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)
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
Bài 1:
a) \(\dfrac{65}{91}+\dfrac{-33}{55}=\dfrac{5}{7}+\dfrac{-3}{5}=\dfrac{25}{35}+\dfrac{-21}{35}=\dfrac{4}{35}\)
b) \(\dfrac{36}{-84}+\dfrac{100}{450}=\dfrac{-3}{7}+\dfrac{2}{9}=\dfrac{-27}{63}+\dfrac{14}{63}=\dfrac{-13}{63}\)
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^^