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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+...+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
ta có 100-(1+1/2+1/3+.....+1/100)
=(1+1+1......1)(99 số 1)-(1+1/2+1/3+......+1/100)
=(1-1)+(1-1/2)+(1-1/3)+.......+(1-1/100)
=1/2+2/3+3/4+.....+99/100
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Leftrightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)
giúp mk vs các bạn ưi ! mk đang cần gấp ai nhanh mik tích cho !nhanh nha help me!thank nhìu
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a, Gọi d là ƯC(12n + 1; 30n + 2 ), ta có :
12n + 1 chia hết cho d => 5( 12n + 1 ) chia hết cho d
30n + 2 chia hết cho d => 2 ( 30n + 2 ) chia hết cho d
-> 5( 12n + 1 ) - 2( 30n + 2 ) chia hết cho d
=> 1 chia hết cho d
vậy d = 1 nên 12n + 1 và 30n + 2 nguyên tố cùng nhau
=> \(\frac{12n+1}{30n+2}\)là phân số tối giản
b, ta có : \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
.....
\(\frac{1}{100^2}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}=\frac{99}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
1/2^2 < 1/(1.2)= 1-1/2
1/3^2 <1/(2.3)=1/2-1/3
1/4^2 <1/(3.4)=1/3-1/4
......
1/100^2 < 1/99-1/100
cộng vế với vế ta được 1/2^2 +1/3^2+...< 1-1/2+1/2-1/3+....+1/99-1/100=1-1/100
=> 100/100-1/100
=>99/100
tk nha bn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\Rightarrow A< 1\)
Vậy A<1
ta có :
\(\frac{1}{2!}=\frac{1}{1.2}\)
\(\frac{1}{3!}=\frac{1}{1.2.3}=\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4!}=\frac{1}{1.2.3.4}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{5!}=\frac{1}{1.2.3.4.5}< \frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
...................................................................................................
\(\frac{1}{99!}=\frac{1}{1.2.3...98.99}< \frac{1}{98.98}=\frac{1}{98}-\frac{1}{99}\)
\(\frac{1}{100!}=\frac{1}{1.2.3....99.100}< \frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)
cộng vế với vế có
\(A=\frac{1}{2!}+\frac{1}{3!}+..+\frac{1}{100!}< \frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\)DPCM