Phần C đề thiếu

\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)

\(\Rightarrow4D=3-\frac{203}{3^{100}}\)

\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)

sửa rồi nhá bn

\(C=\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)

\(\Rightarrow2C=1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)

\(\Rightarrow2C+C=(1-\frac{1}{2}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}})+\)\((\frac{1}{2}-\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}})\)

\(\Rightarrow3C=1-\frac{1}{100}\)

\(\Rightarrow C=\frac{1}{3}-\frac{1}{300}< \frac{1}{3}\left(đpcm\right)\)

Tham khảo nha bạn :

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

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}

\(\frac{1}{2^2}

Đặt tổng sau là B ta có:

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{49^2}+\frac{1}{50^2}\)

Ta lại có :

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(\Rightarrow B< 1-\frac{1}{50}\)

\(\Rightarrow B< 1\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{49^2}+\frac{1}{50^2}\) 

< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{48.49}+\frac{1}{49.50}\)

< \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48.49}+\frac{1}{49.50}=1-\frac{1}{50}

Ta có:

\(\frac{1}{2^2}

ta có :

\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

\(.....................\)

\(\frac{1}{49^2}=\frac{1}{49.49}< \frac{1}{48.49}\)

\(\frac{1}{50^2}=\frac{1}{50.50}< \frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{49^2}+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{48.49}+\frac{1}{49.50}\)

ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{1}{50}\)

\(=\frac{49}{50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{49^2}+\frac{1}{50^2}< \frac{49}{50}\) ( 1 )

mà \(\frac{49}{50}< 1\) ( 2 )

từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+........+\frac{1}{49^2}+\frac{1}{50^2}< 1\)

\(\Rightarrow\text{Đ}PCM\)

Gọi biểu thức trên là A.

Ta có:

A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100

A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100

A < 1 - 1/100

A < 99/100

Mà 99/100 < 1

=> A < 1

đpcm

đúng nhé

gọi A=1/2^2+1/3^2+...+1/50^2

B=1/1.2+1/2.3+...+1/49.50

ta có:

A=1/2^2+1/3^2+...+1/50^2<B=1/1.2+1/2.3+...+1/49.50 (1)

mà B=1/1.2+1/2.3+...+1/49.50

=1-1/2+1/2-1/3+...+1/49-1/50

=1-1/50<1 (2)

kết hợp từ (1) và (2) ta có: A<B<1

=>A<1 (đpcm)

Gọi tổng trên là A

A = 1/2.2 + 1/3.3 +.....+ 1/50.50

A < 1/1.2 + 1/2.3 +.....+ 1/49.50

A < 1 - 1/2 + 1/2 - 1/3 +.......+ 1/49 - 1/50

A < 1 - 1/50

A < 49/50 < 1

=> A < 1 (đpcm)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\Rightarrow A=B\text{(đpcm)}\)

Ta cos ..............

suy ra A=B

bài này chắc mình không làm được rồi, xin lỗihihi

\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)

Vậy \(A< 1\)

Chúc bạn học tốt ~ 

Ta có: \(P=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)

\(\Rightarrow P=\frac{1.2.3....99}{2.3.4...100}\)

\(\Rightarrow P=\frac{1}{100}\)

Ta có: 1/100<1/10 =>P <1/10

nhưng mà bạn ơi, 1/100 làm sao có thể lớn hơn 1/15 được, bạn có sai đề chỗ nào không?