đề bài là Chứng minh hả bạn?????

chỉ cần ns 1 từ dễ

 P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\) 

P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)

P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)

P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)

P\(=\frac{1.51}{50.2}=\frac{51}{100}\)

BÀI 1:

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

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

\(\Leftrightarrow A=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{2^2}\right)+........+\left(\frac{1}{2^{99}+1}+.......+\frac{1}{2^{100}}\right)-\frac{1}{2^{100}}\)

\(\Leftrightarrow A>1+\frac{1}{2}+\frac{1}{2^2}\cdot2+\frac{1}{2^3}\cdot2^2+........+\frac{1}{2^{100}}\cdot2^{99}-\frac{1}{2^{100}}\)

\(\Leftrightarrow A>1+\frac{1}{2}\cdot100-\frac{1}{2^{100}}\)

\(\Leftrightarrow A>51-\frac{1}{2^{100}}>51-1=50\)

\(\Rightarrow DPCM\)

BÀI 2 :

TA CÓ: \(A=1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{100}}\)VÀ \(B=2\)

= > CẦN CHỨNG MINH \(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{100}}\)NHƯ THẾ NÀO SO VỚI 1

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

\(\Leftrightarrow2C=1+\frac{1}{2}+.......+\frac{1}{2^{99}}\)

\(\Leftrightarrow2C-C=\left(1+\frac{1}{2}+.....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+.....+\frac{1}{2^{100}}\right)\)

\(\Leftrightarrow C=1-\frac{1}{2^{100}}>1\)

\(\Rightarrow A>B\)

Bài 2:

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)

Vậy A < 2

Bài 3:

D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)

\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)

Bài 4:

A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)

\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

Ta có:\(-\left(n-1\right)\left(n+1\right)=-\left(n^2+n-n-1\right)=-n^2+1=1-n^2\)

Xét:\(A=...\)

\(=\frac{1-2^2}{2^2}\cdot\frac{1-3^2}{3^2}\cdot...\cdot\frac{1-100^2}{100^2}\)

Mà A có:99 số hạng => số lẻ,nên:

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

\(=-\left(\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{99\cdot101}{100^2}\right)\)

\(=-\left(\frac{1}{2}\cdot\frac{101}{100}\right)=-\frac{101}{200}\)

A=\(\left(\frac{1}{4}-1\right)\)\(\left(\frac{1}{9}-1\right)\)\(\left(\frac{1}{16}-1\right)\)...\(\left(\frac{1}{10000}-1\right)\)

A=\(-\frac{3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-9999}{10000}\)

vì có 50 thừa số nên A= \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)

=>A=\(\frac{3.8.15.....9999}{4.9.16.....10000}\)

A=\(\frac{\left(1.3\right)\left(2.4\right)\left(3.5\right).....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)

A=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)

A=\(\frac{1.101}{100.2}\)

A=\(\frac{101}{200}\)

P = 1/2 + ....1/2^20

2P = 1 + 1/2^2 + 1/2^3+.... + 1/2^19

2P - P 

P = 1-2/2^20

Suy ra P nhỏ hơn 1

= > Chứng tỏ P =.... nhỏ hơn 1 cái này cậu nhập câu hỏi lên google cũng thấy

S thì cậu làm tương tự như P là đc