=1/1*2+1/2*3+...+1/999*1000

=1/1-1/2+1/2-1/3+...+1/999-1/1000

=1-1/1000

So sánh A và B biết;

A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{999}{1000}\)

B = \(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{998}{999}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}+\frac{1}{1000}\)

\(=1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{999}+\frac{1}{999}\right)-\frac{1}{1000}\)

\(=1+0+0+...+0-\frac{1}{1000}\)

\(=1-\frac{1}{1000}=\frac{999}{1000}\)

ngại làm quá

1-1/2+1/3-1/4+......-1/1000 

=(1+1/3+1/5+......+1/999)-(1/2+1/4+.......+1/1000) 

=(1+1/2+1/3+1/4+.....+1/1000)-2(1/2+1/4+.......+1/1000) 

=(1+1/2+1/3+.........+1/1000)-(1+1/2+.....+1/500) 

=1/501 +1/502+1/503+.....+1/1000 ; 

mat khác: 

500-500/501-501/502-.....-999/1000 

=(1-500/501)+(1-501/502)+.....+(1-999/1000)=1/501+1/502+....+1/1000  

=>D=1

\(\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}}{500-\frac{500}{501}-\frac{501}{502}-...-\frac{999}{1000}}=\frac{\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{999}-\frac{1}{1000}\right)}{500-\left(1-\frac{1}{501}\right)-\left(1-\frac{1}{502}\right)-...-\left(1-\frac{1}{1000}\right)}\)

hình như cái mẫu bạn ghi dấu sai thì phải, còn tử thì mình lười làm lắm

tử bạn tính ra 1/2+1/12+...+1/999 000 sau đó phân tích ra là

khó thật

nhớ L-I-K-E nhe tại vì cậu bảo giúp mình, mình cho đúng liền

Ta có :

\(C=\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\)

\(\Rightarrow4C=1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\)

\(\Rightarrow4C-C=\left(1+\frac{1}{4}+.....+\frac{1}{4^{1999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+.....+\frac{1}{4^{1000}}\right)\)

\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{3.4^{1000}}< \frac{1}{3}\)

=> C < 1 / 3

Ta có:

\(C=\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{1000}}\)

\(\Rightarrow4C=1+\frac{1}{4}+...+\frac{1}{4^{999}}\)

\(\Rightarrow4C-C=\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{999}}+\frac{1}{4^{1000}}\right)\)

\(\Rightarrow3C=1-\frac{1}{4^{1000}}\)

\(\Rightarrow C=\left(1-\frac{1}{4^{1000}}\right).\frac{1}{3}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{4^{1000}.3}\)

Mà \(\frac{1}{3}>\frac{1}{3}-\frac{1}{4^{1000}.3}\)

\(\Rightarrow C< \frac{1}{3}\)

Vậy \(C< \frac{1}{3}\)

:)) ko bt làm :))

                                                                                    kí tên

                                                                                   cái nịt

reeeeeeeee

\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)

\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)

\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)

\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)

\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)