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

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 \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\left(đpcm\right)\)

A = 1/2.2 + 1/3.3 + ......+ 1/50.50

A < 1/1.2 + 1/2.3 +......+ 1/49.50

A < 1 - 1/2 + 1/2 - 1/3 +.......+ 1/49 - 1/50

A < 1 - 1/50

A < 49/50 < 1

=> A < 1 (đpcm)

*****k nha

Ta có: A=1/2^2+1/3^2+1/4^2+...+1/50^2<1

=> A<1/1.2+1/2.3+1/3.4+........+1/50.51

=>A< ( 1/1+ -1/2+1/2+ -1/3+1/3+ -1/4+1/4+ -1/5+1/5+.....+1/50+ -1/51)

=> A<1/1+ -1/51

=>A<51/51+ -1/51 =50/51<1

Ta có: 

1/2^2 < 1/1.2

1/3^2 < 1/2.3

......

1/50^2 < 1/49.50

=> A= 1/2^2+1/3^2+...+1/50^2 < 1/1.2+1/2.3+...+1/49.50 = 1-1/2+1/2-1/3+...+1/49-1/50 = 1-1/50<1 (ĐPCM)

Ta có:
A=1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+1/9+......+1/15)+........+ (1/2^99+1/2^99+1+........+1/2^100-1)
(Có 99 nhóm) < 1+2.1/2+2^2.1/2^2+2^3.1/2^3+.....+2^99.1/2^99
=>1+1+1+.......+1 (100 số 1)=100
=>A1+1/2+2.1/2^2+2^2.1/2^3+2^3.1/2^4+.....+2^991/2^100-1-1/2^100 =1+1/2+1/2+1/2+1/2+........+1/2-1/2^100 (100 số 1/2)
=1+100.12-1/2^100
=50+1-1/2^100>50
=>A>50 (2)
Từ (1)và (2)=>50

Ta có: A < \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

Lại có: \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

                                                                                 \(=1+\left(\frac{1}{1}-\frac{1}{50}\right)\)

                                                                                  \(=1+\frac{49}{50}\)

Mà 1+49/50<2 nên A<1+49/50<2

Vậy A<2

1/2^2+1/3^2+...+1/50^2<1/1*2+1/2*3*+...+1/49*50

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

=>S<1+1=2

Vì các p/s bé hơn 1 nên tổng nó bé hơn 1

thế thui

CM: A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\)+...+ \(\dfrac{1}{50^2}\) < 1

      \(\dfrac{1}{2^2}\)  < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)

      \(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)

      .............................

      \(\dfrac{1}{50^2}\) < \(\dfrac{1}{49.50}\) = \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

       Cộng vế với vế ta có:

       A  < \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

       A < 1 - \(\dfrac{1}{50}\)

       A < 1 (đpcm)

ai trả lời nhanh hộ mình nhé