\(M=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{10}{3^{11}}\)

\(\Rightarrow3M=\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\)

\(\Rightarrow3M-M=\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{10}{3^{11}}\right)\)

\(\Rightarrow2M=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}-\frac{10}{3^{11}}\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\)

\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^9}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{10}}< 1\)

\(\Rightarrow2A< 1\)

\(\Rightarrow A< \frac{1}{2}\)

\(\Rightarrow2M< \frac{1}{2}-\frac{10}{3^{11}}\)

\(\Rightarrow M< \frac{\frac{1}{2}-\frac{10}{3^{11}}}{2}\)

\(\Rightarrow M< \frac{1}{4}-\frac{1}{2.3^{11}}< \frac{1}{4}\)

\(\Rightarrow M< \frac{1}{4}\left(đpcm\right)\)

Tham khảo nha bạn :

Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến

a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)

\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)

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

\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)

\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)

\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)

\(\Rightarrow F< \frac{3}{2}\)

\(\Rightarrow2A< 4+\frac{3}{2}\)

\(\Rightarrow2A< \frac{11}{2}\)

\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)

2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)

\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)

\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)

\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)

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

\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)

\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )

\(\Rightarrow2D< 6\)

\(\Rightarrow D< 3\)

\(\Rightarrow2B< 11+3\)

\(\Rightarrow2B< 14\)

\(\Rightarrow B< 7\left(đpcm\right)\)

Phần cuối cũng tương tự 2 phần mình vừa làm nhé

Bạn tự làm nốt nhé đánh mệt lắm

\(\frac{1}{M}=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)

\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)

\(\frac{1}{M}=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{60}\right)\)

\(\frac{1}{M}=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)

-theo t đề là M chứ ko phải 1/M 

Ta có : \(\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\)

\(\Rightarrow m=\frac{30}{19}>\frac{2}{3}\)

\(Tac\text{ó}:\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{59}-\frac{1}{60}\right)\)

\(=>2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\\ =>m=\frac{30}{19}>\frac{2}{3}\)

M = 1/3 + 2/3² + 3/3³ + 4/3^4 + ... + 100/3^100 

=> 3M= 1 + 2/3 + 3/3² + 4/3³ + .... + 100/3^99

=> 3M-M = 1 + ﴾2/3 ‐ 1/3﴿ + ﴾3/3² ‐ 2/3²﴿ +...+ ﴾100/3^99 ‐ 99/3^99﴿ ‐ 100/3^100 

=> 2M= 1+ 1/3 + 1/3² + 1/3³ +...+ 1/3^99 ‐ 100/3^100 

Đặt N = 1/3 + 1/3² + 1/3³ +...+ 1/3^99

=> 3N = 1 + 1/3 + 1/3² + 1/3³ +...+ 1/3^98

=> 2N = 1 ‐ 1/3^99 

=> N = ﴾1 ‐ 1/3^99﴿/2 

Thay vào 2M 

=> 2M= 1+ 1/2 ‐ 1/﴾2x3^99﴿ ‐ 100/3^100 < 1+ 1/2 = 3/2 

=> M < 3/4 

vậy...

Bài này công nhận là dễ , nhưng khi nãy bận ăn cơm ,  xin lỗi ha!! Hứa lần sau sẽ giải cho!!!

Nguyễn Ngọc Quý ukm, lần sau tui đăng bài khác lên cho mà giải

\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{15}{14}>1\left(1\right)\)

\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}

\(\frac{1}{M}=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+...+\frac{1}{1+2+3+...+59}\)

\(\frac{1}{M}=\frac{1}{3\left(1+3\right):2}+\frac{1}{4\left(1+4\right):2}+\frac{1}{5\left(1+5\right):2}+...+\frac{1}{59\left(1+59\right):2}\)

\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)

\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{60}\right)\)

\(\frac{1}{M}=\frac{1}{2}.\frac{19}{60}\)

\(\frac{1}{M}=\frac{19}{120}\)

\(M=\frac{120}{19}>\frac{2}{3}\left(đpcm\right)\)

chuẩn men