Chứng minh :
\(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ ... + \(\frac{1}{50^2}\) < 1
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Những câu hỏi liên quan
PT
4
LB
6 tháng 5 2017
\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
V
1
DL
5 tháng 5 2017
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(\frac{1}{1^2}=1\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\left(dpcm\right)\)
HP
CHO \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}.\)CHỨNG MINH A<2
4
11 tháng 4 2017
\(\frac{1}{2^2}< \frac{1}{1.2}\)
...................\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)
10 tháng 4 2017
1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50
ta có:
=> 1/1^2+1/2*3+1/3*4+...+1/49*50
<=> 1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
<=> 1-1/50 < 2
=> A < 2
17 tháng 3 2017
TA CÓ Vế trái <\(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
do đó VT <2(dpcm)
2 tháng 5 2017
\(A=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
Khi đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}=B\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{49}-\frac{1}{50}< 1\)
Vì \(A< 1+B\)mà \(B< 1\)nên \(B+1< 2\)do đó \(A< 2\)
Vậy \(A< 2\)
2 tháng 5 2017
1/12+1/22+....+1/502<1/1+1/1x2+1/2x3+....+1/49x50=1-1/50=49/50<2
=>A<2(đpcm)
TN
8 tháng 4 2016
đặt B=1/2.3+1/3.4+...+1/49.50
=1/1.2+1/2.3+1/3.4+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/50.50<1-1/2+1/2-1/3+...+1/49-1/50 (3)
từ (1),(2),(3) =>A<2
BM
8 tháng 4 2016
Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}=1+\frac{1}{2^2}+........+\frac{1}{50^2}\)
=> \(A<1+\frac{1}{1.2}+\frac{1}{2.3}+.............+\frac{1}{49.50}\)
=> \(A<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{49}-\frac{1}{50}\)
=> \(A<2-\frac{1}{50}\Rightarrow A<2\)
Vậy A nhỏ hơn 2
23 tháng 4 2016
Ta có
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3}......\frac{1}{50^2}<\frac{1}{49.50}\)
\(=>A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=> A<2-1/50
=> A < 2
=> đpcm
23 tháng 4 2016
Ta có: A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
A < \(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=> A < 1 +( \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\))
A< 1 +1 -\(\frac{1}{50}\)
A< 2 - \(\frac{1}{50}\)
Vậy A< 2
Ta có:\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{50^2}\)<\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+...+\(\frac{1}{49\cdot50}\)
<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
<1-\(\frac{1}{50}\)<1
Nên \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{50^2}\)<1
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}=S\)
Đặt S = \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
Ta lại có: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};....;\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{50}=\frac{49}{50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< S=\frac{49}{50}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\) (đpcm)