m=1/51+1/52+.......+1/100 / 1/1*2 + 1/2*3 + ........+1/99*100
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
NT
1
24 tháng 4 2018
Ta có \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(\Rightarrow\text{Đ}PCM\)
L
1
11 tháng 4 2023
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
NV
0
NT
2
S
25 tháng 7 2016
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)= \(\left(1+\frac{1}{3}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}+\frac{1}{100}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
\(\frac{1}{51}+\frac{1}{52}+....+\frac{1}{100}=-\frac{1}{2}\)
KM
1
24 tháng 6 2018
Đặt \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow A-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)
\(\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)
BN
0
CN
2
TN
21 tháng 5 2016
= (1+1/3+1/5+…+1/99)-(1/2+1/4+….+1/100)
= (1+1/2+1/3+…+1/100)-2(1/2+1/4+1/6+…+1/100)
= (1+1/2+1/3+…+1/100)-(1+1/2+1/3+…+1/50)
=1/51+1/52+…+1/100=VP (đpcm)
21 tháng 5 2016
= (1+1/3+1/5+…+1/99)-(1/2+1/4+….+1/100)
= (1+1/2+1/3+…+1/100)-2(1/2+1/4+1/6+…+1/100)
= (1+1/2+1/3+…+1/100)-(1+1/2+1/3+…+1/50)
=1/51+1/52+…+1/100=VP (đpcm)