Cho:A = 1/(1·2^2) + 1/(2·3^2) + 1/(3·4^2) + ... + 1/(49·50^2)B = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/50^2Chứng minh rằng: A < 1/2 < B help me help me help me
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CM
1
DA
17 tháng 6 2020
1/2 + 1/2^2 + 1/3^2 + .....+ 1/50^2 < 1/1 + 1/1.2 + 1/2.3 +...+ 1/49.50
Đặt A = 1/1 + 1/1.2 + 1/2.3 +...+ 1/49.50
A= 1/1 - 1/1 + 1/1 -1/2 + 1/2 -1/3+...+ 1/49-1/50
A= 1/1 - 1/50
A= 49/50
Vì 49/50 < 1 mà 1/2 + 1/2^2 + 1/3^2 + .....+ 1/50^2 < 49/50 nên 1/2 + 1/2^2 + 1/3^2 + .....+ 1/50^2 <1
Vậy....
NT
1
22 tháng 4 2016
Ta có : 1/2^2<1/1.2
1/3^2 < 1/2.3
1/4^2<1/3.4
................
.............
1/50^2<1/49.50
=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 1/1.2+1/2.3+1/3.4+....+1/49.50
=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 1-1/50
=> 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 49/50 < 1
Vậy 1/2^2+1/3^2+1/4^2+1/5^2+.....+1/50^2 < 1
NQ
0
TT
1
4 tháng 4 2017
Đặt A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
Với n \(\in\) N*, n > 1 ta có :
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)( vì 1>0; n2 > n(n-1) > 0 )
Áp dụng vào bài ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
.....
\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
=> \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{50^2}\)< \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
=> A < \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)
=> A < \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\)
=> A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
=> A < \(1-\dfrac{1}{50}\) < 1 ( vì \(\dfrac{1}{50}>0\) )
=> A < 1
=> đpcm
Vậy...
NH
0
T
0
LK
2
17 tháng 5 2016
Đặt A=1/2^2+1/3^2+1/4^2+...+1/50^2
A<1/1*2+1/2*3+1/3*4+...+1/49*50
A<1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
A<1-1/50<1
Vậy A<1
BH
17 tháng 5 2016
Ta có:\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\left(đpcm\right)\)
HT
1
20 tháng 4 2017
\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)<1
ta có \(\frac{1}{2^2}\)<\(\frac{1}{1.2}\)
\(\frac{1}{3^2}\)<\(\frac{1}{2.3}\)
..........................
\(\frac{1}{50^2}\)<\(\frac{1}{49.50}\)
ta được \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{49.50}\)
=>1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-...-\(\frac{1}{49}\)+\(\frac{1}{49}\)-\(\frac{1}{50}\)
=>1-\(\frac{1}{50}\)<1 nên\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)<1
vậy ...........................
TQ
2
B
9 tháng 5 2016
1/22 < 1/2.3 ; 1/32 < 1/3.4 ; .....; 1/502 < 1/50.51 => A < 1+1-1/2+1/2-1/3+...1/50-1/51 < 2