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S=1/50+1/51+...+1/98+1/99
Ta thấy :
1/50>1/100
1/51>1/100
................
1/99>1/100
1/100=1/100
=>1/51+1/52+1/53+...+1/98+1/99>1/100+1/100+1/100+...+1/100 (Mỗi bên 50 số hạng)
=>S>50.1/100
=>S>50/100=1/2
Vậy S>1/2
Đặt S=1/4+1/16+1/36+...+1/10000
S= 1/4x(1+1/4+1/9+...+1/2500)
S= 1/4x(1+1/2x2+1/3x3+...+1/50x50)
S< 1/4x(1+1/1x2+1/2x3+...1/49x50)
S< 1/4x(1+1-1/2+1/2-1/3+....+1/49-1/50)
S< 1/4x(1+1-1/50)
S< 1/4x(2-1/50)<2/4(2/4=1/2)
S< 1/2
Ta có: \(\frac{1}{4}< \frac{1}{2}\)
\(\frac{1}{16}< \frac{1}{2}\)
... . . .
\(\frac{1}{10000}< \frac{1}{2}\)
\(\frac{1}{10000}+\frac{1}{10000}+...+\frac{1}{10000}< \frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{10000}< \frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\)(*) (n phân số \(\frac{1}{10000}\) ; n phân số \(\frac{1}{2}\))
Từ đó suy ra \(\frac{1}{4}+\frac{1}{6}+\frac{1}{36}+\frac{1}{64}+...+\frac{1}{1000}< \frac{1}{2}\left(đpcm\right)\)
Đặt \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+....+\frac{1}{17}\)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{6}{5}\left(1\right)\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{7}{11}\left(2\right)\)
Từ (1)(2) \(\Rightarrow A< \frac{6}{5}+\frac{7}{11}=\frac{66}{55}+\frac{35}{55}=\frac{101}{55}< \frac{110}{55}=2\)
\(\Rightarrow A< 2\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 2\left(đpcm\right)\)
Ta có:
1/5=1/5
1/6<1/5
1/7<1/5
..........
1/10<1/5
=>1/5+1/6+...+1/10<1/5.6=6/5(1)
Lại có :
1/11=1/11
1/12<1/11
1/13<1/11
.............
1/17<1/11
=>1/11+1/12+1/13+...+1/17<1/11.7=7/11(2)
Từ (1)và (2)=>1/5+1/6+...+1/17<6/5+7/11=101/55<110/55=2
=>1/5+1/6+...+1/17<2
ĐPCM
Ta có\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\)<\(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}\)=\(\frac{5}{6}\)(6 c/s \(\frac{1}{5}\))
Ta lại có \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{17}\)<\(\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\)=\(\frac{7}{11}\)(7 c/s \(\frac{1}{11}\))
Suy ra \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)<\(\frac{110}{55}\)=2
Vậy...
Hok tốt
Đặt \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{6}{5}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{7}{11}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< \frac{6}{5}+\frac{7}{11}\)
\(\Rightarrow A< \frac{101}{55}< \frac{110}{55}=2\)
\(\Rightarrow A< 2\)( ĐPCM )
1)Tính
S=1+2+3+.......+(m-1)
2)cho a nguyên tố lớn hơn 3 chứng tỏ :
A=2+(a+2)(a+1) chia ht cho 12
1.
a.Để A là phân số thì n - 5 khác 0 => n khác 5
b.Để A \(\in\)Z thì 3 chia hết cho n - 5 => n - 5 \(\in\) Ư(3) = {1; 3; -1; -3}
Ta có bảng sau:
| n - 5 | 1 | -1 | 3 | -3 |
| n | 6 | 4 | 8 | 2 |
Vậy n \(\in\){6; 4; 8; 2} thì A \(\in\)Z.