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Thêm bớt ở A phân số 1/2¹⁰⁰

\(A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{2^2}\right)+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2^3}\right)+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{99}+1}+...+\frac{1}{2^{100}-1}+\frac{1}{2^{100}}\right)+\frac{1}{2^{100}}\)

\(\Rightarrow A\ge1+\frac{1}{2}+\frac{2}{2^2}+\frac{4}{2^3}+\frac{8}{2^4}+...+\frac{2^{99}}{2^{100}}-\frac{1}{2^{100}}=1+\frac{1}{2}+...+\frac{1}{2}-\frac{1}{2^{100}}\)( 100 ps 1/2)\(\Rightarrow A>1+50-\frac{1}{2^{100}}>50\)

=> ĐPCM

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}< 1\)

Vậy \(A< 1\left(đpcm\right)\)

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

\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{50}\)

\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}\)

\(\Leftrightarrow B< \frac{3}{4}\left(đpcm\right)\)

\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}...+\frac{1}{100!}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\Leftrightarrow C< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Leftrightarrow C< 2-\frac{1}{100}\)

\(\Leftrightarrow C< 2\left(đpcm\right)\)

Gọi biểu thức trên là A.

Chứng minh A > 50

\(A=1+\frac{1}{2}+\left(\frac{1}{2^1+1}+\frac{1}{2^2}\right)+\left(\frac{1}{2^2+1}+\frac{1}{6}+...+\frac{1}{2^3}\right)+...+\left(\frac{1}{^{2^{100-2}+1}}+...+\frac{1}{2^{100-1}}\right)\\ \)

\(A>1+\frac{1}{2}+\frac{1}{2^2}.2+\frac{1}{2^3}.2^2+...+\frac{1}{2^{100-1}}2^{100-2}\)

\(A>\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{2}\)

\(< =>A>\frac{100}{2}=50\)

Chứng minh A<100

\(A=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{2^2}+\frac{1}{5}+...+\frac{1}{7}\right)+....+\left(\frac{1}{2^{100-2}}+\frac{1}{2^{100-2}+1}+...+\frac{1}{2^{100-1}-1}\right)\)-\(\frac{1}{2^{100-1}}\)

\(A< 1+\frac{1}{2}.2+\frac{1}{2^2}.2^2+...+\frac{1}{2^{100-2}}.2^{100-2}+\frac{1}{2^{100-1}}\)

\(A< 1+1+1+...+1+\frac{1}{2^{100-1}}\)

\(A< 1.99+\frac{1}{2^{100-1}}< 99+1=100\)

ta có : 1+1/2+1/3+....+1/2^100-1   

= 1/2x2 +1/3x2 +1/4x2 +...+ 1/2^100 x2

= 2x(1/2+1/3+1/4+...+1/2^100)      

=.................... làm đến đây mk tịt

mk có chacha

15135454

Ta có: \(55+5\)1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 +.....+ 1/50^2  =  1/1^2 + 1/2^2 + (1/3^2 + 1/4^2 +....+ 1/50^2 )

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

                                                                               = 1,73 = 173/100 (dpcm) 

\(VP=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(VP=\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)

\(VP=\frac{2}{2}-\frac{1}{2}+\frac{3}{3}-\frac{1}{3}+\frac{4}{4}-\frac{1}{4}+...+\frac{100}{100}-\frac{1}{100}\)

\(VP=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)

\(VP=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=VT\) ( đpcm ) 

Mk nghĩ \(VT=100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\) bn xem lại đề có nhầm ko 

Chúc bạn học tốt ~ 

ko mk thấy đúng mà

ko nhầm đề đâu