\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}

Ta có : 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

= 1 - 1/50 < 1

Nên 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50 < 1

A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{5.6}\)+....+ \(\dfrac{1}{49.50}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+ \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

A = 1 - \(\dfrac{1}{50}\) < 1

A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{3.4}\)+.....+ \(\dfrac{1}{49.50}\) < 1 ( đpcm)

a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦ 
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) = 
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 < 
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥ 
♦, ♥ => 7 / 12 < A < 5 / 6

b)ta có:

1/1.2+1/3.4+1/5.6+...+1/49.50

=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)

=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2

=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)

=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50

hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50

đặt A=1/1.2+1/2.3+1/3.4+..........1/49.50

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=1-\frac{1}{50}<1\)

vậy A<1

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50

1 - 1/50 < 1

1/1.2 + 1/2.3 + 1/3.4 + ...... + 1/49.50

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... + 1/49 - 1/50

1 - 1/50 < 1

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

=>\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

=>\(A=1-\frac{1}{50}=\frac{49}{50}\)

mà A=49/50 

=>1/26+1/27+...+1/50 =49/50

49/50 ban oi

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