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ta co 3+1/3+9+1/9+27+1/27+...

minh quen roi vao trang cua quan thanh thot nhe!

= \(1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)\(\frac{1}{3^{98}}\)

\(=1.98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

Đặt A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

\(\Rightarrow3A-A=2A=1-\frac{1}{3^{98}}\Rightarrow A=\frac{1-\frac{1}{2^{98}}}{2}< 1\)

\(\Rightarrow B=98+A< 98+1< 99< 100\)

\(\Rightarrow B< 100\)

Xét \(B=\frac{4}{3}+\frac{10}{9}+...+\frac{3^{98}+1}{3^{98}}\)

   \(\Leftrightarrow B=\frac{3+1}{3}+\frac{9+1}{9}+...+\frac{3^{98}+1}{3^{98}}\)

   \(\Leftrightarrow B=\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{1}{3^{98}}\right)\)(có 98 cặp số hạng)

\(\Leftrightarrow B=\left(1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)(có 98 số hạng 1)

\(\Leftrightarrow B=98+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

Lấy 3A-A, ta được:

\(2A=1-\frac{1}{3^{98}}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2\cdot3^{98}}\)(*)

Thay (*) vào biểu thức B, ta được

\(B=98+\frac{1}{2}-\frac{1}{2\cdot3^{98}}< 100\)

VẬY, B<100 (ĐPCM)

Ta có : 

\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}\)

\(B=\frac{3+1}{3}+\frac{9+1}{9}+\frac{27+1}{27}+...+\frac{3^{98}+1}{3^{98}}\)

\(B=\frac{3}{3}+\frac{1}{3}+\frac{9}{9}+\frac{1}{9}+\frac{27}{27}+\frac{1}{27}+...+\frac{3^{98}}{3^{98}}+\frac{1}{3^{98}}\)

\(B=1+\frac{1}{3}+1+\frac{1}{9}+1+\frac{1}{27}+...+1+\frac{1}{3^{98}}\)

\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^{98}}\right)\)

\(B=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

Do từ \(1\) đến \(98\) có \(98-1+1=98\) số hạng nên có \(98\) số \(1\) suy ra : 

\(B=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\) ta có : 

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)\)

\(2A=1-\frac{1}{3^{98}}< 1\)

Mà \(2A< 1\)\(\Rightarrow\)\(A< 1\)

Do đó : 

\(B=98+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)< 98+1=99< 100\)

\(\Rightarrow\)\(B< 100\) ( đpcm ) 

Vậy \(B< 100\)

Chúc bạn học tốt ~