a) dat A=1+2+2²+2³+...+2⁹⁹

2.A=2+2²+2³+2⁴+...+2¹⁰⁰

2.A-A= 2+2³+2⁴+...+2¹⁰⁰-(1+2+2²+2³+...+2⁹⁹)

A=2¹⁰⁰-1

----> 1.3.5.7...197.199<\(\frac{101.102.103....200}{2^{100}-1}\)

Dat B =1.3.5.7...197.199 

B=\(\frac{1.3.5.7....197.199...2.4.6.8....200}{2.4.6.8....200}\)

B= \(\frac{1.2.3.4.5....199.200}{2.4.6.8....200}\)

B=\(\frac{1.2.3.4.5......199.200}{2^{100}.\left(1.2.3.4...100\right)}\) ( tu 2 den 200 co 100 so hang nen duoc 2¹⁰⁰)

B =\(\frac{101.102.103....200}{2^{100}}\)

---->\(\frac{101.102.103....200}{2^{100}}

b> A= \(\frac{1.3.5.7....2499}{2.4.6.8....2500}\)  chon B=\(\frac{2.4.6.8...2500}{3.5.7.9...2501}\)

A.B = \(\frac{1.3.5.7....2499.2.4.6.8...2500}{2.4.6.8...2500.3.5.7.....2499.2501}=\frac{1}{2501}\)

Nhan xet 

\(\frac{1}{2}+\frac{1}{2}=1\)

\(\frac{2}{3}+\frac{1}{3}=1\)

vi 1/2 >1/3----> 1/2 <2/3

cm tuong tu ta se co A<B

---> A.A<A.B

---->A²<A.B

===> A² <\(\frac{1}{2501}

a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

. . .

\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)

b) Ta có :

\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)

\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)

Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

. . .

\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)

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

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

\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)

\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)

hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)

a) gọi \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

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

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

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

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

\(=2-\frac{1}{50}< 2\)

\(\Rightarrow A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)

b) Ta thấy \(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\), \(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)

Do đó : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}.3+\frac{1}{45}.3=\frac{1}{2}\)

c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)

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

Ta thấy vế trong ngoặc nhỏ hơn 1

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>48\)

A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

A = 1 - 1/100

A = 99/100

B = 5/1*4 + 5/4*7 + .... + 5/100*103

B = 5/3*(3/1*4 + 3/4*7 + ... + 3/100*103)

B = 5/3*(1 -1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)

B = 5/3*(1 - 1/103)

B = 5/3* 102/103

gọi ƯC(n + 1; n + 2) = d

=> n + 1 chia hết cho d và n + 2 chia hết cho d

=> n + 2 - n - 1 chia hết cho d

=> 1 chia hết cho d

=> d = + 1

=> n+1/n+2 là phân số tối giản với mọi n là stn

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

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

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

\(A=\frac{1}{2}-\frac{1}{100}\)

\(A=\frac{49}{100}\)