\(CTR:1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{200}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
7 tháng 4 2017
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{100}\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)+\left(\frac{1}{101}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{199}+\frac{1}{200}\) (ĐPCM)
7 tháng 4 2017
Ta có : 1 - 1/2 + 1/3 - 1/4 + ....- 1/200
= (1 + 1/3 + 1/5 + ....+ 1/199) - ( 1/2 + 1/4 + 1/6 + .... + 1/200)
= ( 1 + 1/3 +...+ 1/199) + (1/2 +1/4 + ...+ 1/200) - 2(1/2+1/4+...+ 1/200)
= (1+1/2+1/3+....+1/199 + 1/200) - (1 +1/2 +1/3 +....+1/100)
= 1/101 + 1/102+ 1/103 + .... + 1/200
chúc bạn học tốt!!!!!!!
1 tháng 4 2016
Ta có : \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)\(\left(đpcm\right)\)
KN
13 tháng 4 2019
Tham khảo ở link này bạn nhé :
https://olm.vn/hoi-dap/detail/5631756599.html
~ Study well ~
27 tháng 2 2018
Biến đổi vế trái ta có :
\(VT=\frac{1}{1}+\frac{1}{3}+...+\frac{1}{199}+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)-\) \(2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}+...+\frac{1}{200}-\) \(1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\) \(=VP\RightarrowĐPCM\)
19 tháng 5 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
AK
19 tháng 5 2018
Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\left(đpcm\right)\)
Chúc bạn học tốt !!!
AH
Akai Haruma
Giáo viên
22 tháng 2 2020
Lời giải:
Ta có:
\(\text{VT}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\text{VP}\)
Ta có đpcm.
18 tháng 9 2019
Biến đổi vế phải của đẳng thức :
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{100}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right]\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)
29 tháng 2 2016
1-1/2+1/3-1/4+...+1/199-1/200
=(1+1/3+...+1/199)-(1/2+1/4+...+1/200)
=(1+1/2+1/3+...+1/199+1/200)-2(1/2+1/4+...+1/200)
=(1+1/2+1/3+...+1/199+1/200)-(1+1/2+...+1/100)
=1/101+1/102+...+1/200 (đpcm)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Vậy \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{200}\)