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Bài 1:
C = 1/101 + 1/102 + 1/103 + ... + 1/200
Có:
C < 1/101 + 1/101 + 1/101 + ... + 1/101
C < 100 . 1/101
C < 100/101
Mà 100/101 < 1
=> C < 1 (1)
Có:
C > 1/200 + 1/200 + 1/200 + ... + 1/200
C > 100 . 1/200
C > 1/2 (2)
Từ (1) và (2)
=> 1/2<C<1
Ủng hộ nha mk làm tiếp
Ta có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{1}{300}.200=\dfrac{200}{300}=\dfrac{2}{3}\)
\(\Rightarrow\) biểu thức trên lớn hơn \(\dfrac{2}{3}\).
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\)
\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)
\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{10}\)
\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)
\(\Leftrightarrow D< 1\left(đpcm\right)\)
a, 52015+52014+52013 chia hết cho 31
52015+52014+52013
=52013.(52+5+1)
=52013.31
Vì 31 chia hết cho 31
=> 52013.31 chia hết cho 31
Hay 52015+52014+52013 chia hết cho 31.
b, 439+440+441 chia hết cho 28
439+440+441
=438.(4+42+43)
=438.84
Vì 84 chia hết cho 28
=> 438.84 chia hết cho 28
Hay 439+440+441 chia hết cho 28.
c, 1+7+72+.....+7101 chia hết cho 8
1+7+72+.....+7101
=(1+7)+72.(1+7)+....+7100.(1+7)
=8+72.8+....+7100.8
=8(1+72+....+7100)
Vì 8 chia hết cho 8
=> 8(1+72+....+7100) chia hết cho 8
Hay 1+7+72+.....+7101 chia hết cho 8.
a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
Câu 8( Mình không viết đè nữa nha)
a) 2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100
= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100
= 1 – 1/100 < 1
= 99/100 < 1
Vậy A< 1
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>200.\frac{1}{300}\)
\(>\frac{2}{3}\)
Ta có
\(\frac{1}{101}>\frac{2}{3}\)
\(\frac{1}{102}>\frac{2}{3}\)
.
.
.
\(\frac{1}{300}>\frac{2}{3}\)
Vậy \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
a) \(2\cdot\left(1+2+2^2\right)=2\cdot7⋮7\left(đpcm\right)\)
b) \(3\cdot\left(1+3+3^2\right)=3\cdot13⋮13\left(đpcm\right)\)
Bài 1 : Chứng tỏ rằng :
a) 2 + 2^2 + 2^3
= 2 . 1 + 2^1 . 2^1 + 2^1 . 2^2
= 2 . 1 + 2 . 2 + 2 . 4
= 2 . ( 1 + 2 + 4 )
= 2 . 7 chia hết cho 7 .
Vậy 2 + 2^2 + 2^3 chia hết cho 7
b) 3^100 + 3^101 + 3^102
= 3^100 . 1 + 3^100 . 3^1 + 3^100 . 3^2
= 3^100 . 1 + 3^100 . 3 + 3^100 . 9
= 3^100 . ( 1 + 3 + 9 )
= 3^100 . ( 4 + 9 )
= 3^100 . 13 chia hết cho 13
Vậy 3^100 + 3^101 + 3^102 chia hết cho 13 .