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

thế thui

5 tháng 11 2023

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)

 

Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)(đpcm)

28 tháng 6 2021

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)   (đpcm)

9 tháng 10 2016

Ta biến đổi vế phải :

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

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

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

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

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

Vậy 1/26+1/27+1/28+.....+1/50=1-1/2+1/3-1/4+......+1/49-1/50

Mình không bấm phân số được mong mấy bạn thông cảm

 1/26+1/27+1/28+...+1/49+1/50=1-1/2+1/3-1... 
<=>2/26+2/28+2/30+...+2/50=1-1/2+1/3-1... 
<=>1/13+1/14+1/15+...+1/25=1-1/2+1/3-1... 
<=>2/14+2/16+2/18+...2/24=1-1/2+1/3-1/... 
<=>1/7+1/8+1/9+...+1/12=1-1/2+1/3-1/4+... 
<=>2/8+2/10+2/12=1-1/2+1/3-1/4+1/5-1/6 
<=>1/4+1/5+1/6=1-1/2+1/3-1/4+1/5-1/6 
<=>2/4+2/6=1-1/2+1/3 
<=>1/2+1/3=1-1/2+1/3 
<=>2/2=1

 

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

=>đpcm

12 tháng 5 2023

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

17 tháng 10 2018

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

\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{40}+\frac{1}{50}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)

Vậy .....(tự kết luận)

17 tháng 10 2018

CMR: \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

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

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}^{\left(đpcm\right)}\)

9 tháng 10 2016

Ta có :1/26 + 1/27 + ... + 1/50 - (1-1/2+1/3-1/4+...+1/49-1/50) 
=1/26+1/27+...+1/50 + (1/26-1/27+....-1/49+1/50) + (-1/13+1/14-....+1/24-1/25)+(-1/7+1/8-..... + 1/12) + (1/6-1/5+1/4)+(1/2-1) 
=1/13+1/14+...+1/25+ (-1/13+1/14-....+1/24-1/25)+(-1/7+1/8-..... + 1/12) + (1/6-1/5+1/4)+(1/2-1) 
=1/7+1/8+...+1/12 + (-1/7+1/8-...-1/11 + 1/12) + (1/6-1/5+1/4)+(1/2-1) 
=1/4+1/5+1/6 +(1/6-1/5+1/4)+(1/2-1) 
=1/2+1/2-1 
=0 
Vậy 1/26 + 1/27 + 1/28 +.....+ 1/49 +1/50 = 1- 1/2 +1/3 - 1/4 +....+ 1/49 - 1/50

5 tháng 5 2017

gọi \(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)và \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

Ta có : \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

         \(B=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

         \(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

         \(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

        \(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=A\)

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

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

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

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