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H
3
28 tháng 6 2021
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)
SI
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)
LT
1
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
PN
9
28 tháng 7 2015
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
28 tháng 7 2015
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
=>đpcm
KH
1
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
BN
1
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
DH
1
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 : \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=A\)
HT
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
0