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DK
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0
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0
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NT
2
AK
11 tháng 3 2018
\(S=\frac{5}{2^2}+\frac{5}{3^2}+\frac{5}{4^2}+...+\frac{5}{100^2}\)
\(S=5.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Ta có : \(\frac{1}{2^2}>\frac{1}{2.3},\frac{1}{3^2}>\frac{1}{3.4},\frac{1}{4^2}>\frac{1}{4.5},...,\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(\Rightarrow5.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)>5.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(\Rightarrow S>5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow S>5.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow S>5.\frac{99}{202}\)
\(\Rightarrow S>\frac{495}{202}>\frac{404}{202}=2\)
\(\Rightarrow S>2\)
AK
11 tháng 3 2018
\(CM:S< 5\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2},\frac{1}{3^2}< \frac{1}{2.3},...,\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}\)
\(\Rightarrow5.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)< 5.\frac{99}{100}\)
\(\Rightarrow S< \frac{495}{100}< \frac{500}{100}\)
\(\Rightarrow S< 5\)
ND
1
7 tháng 4 2019
5/22 + 5/32 + 5/42 +...+ 5/1002 < 5/1.2 + 5/2.3 +5/3.4 +...+ 5/99.100
5/2.2 +5/3.3 + 5/4.4 +...+ 5/100.100 < 5. ( 1/1.2 + 1/2.3 +1/3.4 +..+ 1/99.100)
5/2.2 +5/3.3 + 5/4.4 +...+ 5/100.100 < 5. (1/1 -1/2 +1/2 -1/3 +1/3-1/4 +...+ 1/99-1/100)
5/2.2 +5/3.3 + 5/4.4 +...+ 5/100.100 < 5. (1/1-1/100)
5/2.2 +5/3.3 + 5/4.4 +...+ 5/100.100 < 5. ( 100/100 -1/100)
5/2.2 +5/3.3 + 5/4.4 +...+ 5/100.100 < 5. 99/100
5/2.2 +5/3.3 + 5/4.4 +...+ 5/100.100 < 99/20
mình chỉ giải tới đây thôi vì đã dễ rồi
Cho: \(A=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+....+\dfrac{2}{100^2}\)
\(A=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)
Và cho \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
Mà:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
....
\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)
Nên: \(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow B< 1-\dfrac{1}{100}\)
\(\Rightarrow B< \dfrac{99}{100}\)
Mà: \(\dfrac{99}{100}< 1\) (tử nhỏ hơn mẫu)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
\(\Rightarrow A=2\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{100^2}\right)< 2\) (đpcm)
\(\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+...+\dfrac{2}{100^2}\)
\(=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)
mà \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
\(\Rightarrow dpcm\)