ko ai trả lời thì để mình

C/M : n/n+1 < n+1/n+2

1 - n/n+1 = 1/n+1

1 - n/n + 2 = 1/n+2

Vì 1/n+1 > 1/n+2 nên n/n+1 < n+1/n+2

1/2 . 3/4 . 5/6 ... 2499/2500 < 1/2 . 2/3 . 3/4 ... 2501/2502

=1/2501 < 1/2500 (1/50) ²

1/50 < 1/49 => A <1/49

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

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

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

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

=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)

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

(: ko bít. tui giỏi tiếng anh nhưng ngu toán lắm

b: \(B=\frac34+\frac89+\frac{15}{16}+\cdots+\frac{2499}{2500}\)

\(=1-\frac14+1-\frac19+\cdots+1-\frac{1}{2500}\)

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

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

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

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}<1\)

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

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

=>B>48

=>48<B<49

=>B không là số nguyên

a: \(A=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\cdots+\frac{21}{10^2\cdot11^2}\)

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

\(=1-\frac{1}{11^2}\)

=>A<1