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

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

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

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

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

nên B>A

A là số nào vậy bạn giải thích rõ giùm

B = 3/4 + 8/9 + 15/16 + .... + 2499/2500

B = (1 - 1/4) + (1 - 1/9) + (1 - 1/16) + ... + (1 - 1/2500)

B = (1 - 1/2²) + (1 - 1/3²) + (1 - 1/4²) + ... + (1 - 1/50²)

B = (1 + 1 + 1 + ... + 1) - (1/2² + 1/3² + 1/4² + ...+ 1/50²)

                49 số 1

B = 49 - (1/2² + 1/3² + 1/4² + ... + 1/50²)

=> B < 49 (1)

B > 49 - (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/49×50)

B > 49 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50)

B > 49 - (1 - 1/50)

B > 49 - 1 + 1/50

B > 48 + 1/50 > 48 (2)

Từ (1) và (2) => 48 < B < 49

=> B không phải là số nguyên ( đpcm)

B = 3/4 + 8/9+ 15/16 + ... + 2499/2500

B = (1 - 1/4) + (1 - 1/9) + (1 - 1/16) + ... + (1 - 1/2500)

B = (1 - 1/2²) + (1 - 1/3²) + (1 - 1/4²) + ... + (1 - 1/50²)

B = (1 + 1 + 1 + ... + 1) - (1/2² + 1/3² + 1/4² + .... + 1/50²)

              49 số 1

=> B = 49 - (1/2² + 1/3² + 1/4² + ... + 1/50²)

=> B < 49 (1)

B > 49 - (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/49×50)

B > 49 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50)

B > 49 - (1 - 1/50)

B > 49 - 1 + 1/50

B > 48 + 1/50 > 48 (2)

Từ (1) và (2) => 48 < M < 49

=> M không phải số nguyên ( đpcm)

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

ko ngờ đấy mày lại ko được giải khi thi MYTS
 

MYTS  là j ạ

Sai đề à bạn H>49 thì đc

\(H=2+\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

\(=2+1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{2500}\)

\(=2+49-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)

\(=51-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}\right)\)

Do \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};...;\frac{1}{50.50}< \frac{1}{49.50}\)

Nên \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

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

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

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

\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}< 1\)

\(\Rightarrow H=51-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}\right)>51-1=50\)

Vậy H>50