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

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

A = \(\frac{1.3.2.4.3.5.......49.51}{2.2.3.3.4.4........50.50}\)

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

A = \(\frac{51}{50.2}\)

A = \(\frac{51}{100}\)

A = \(\frac{1.3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\cdot\cdot\frac{49\cdot51}{50\cdot50}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot49\cdot51}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot\cdot\cdot\cdot50\cdot51}=\frac{1\cdot51}{2\cdot50}=\frac{51}{100}\)

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

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

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

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

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

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{50^2}< \frac{1}{49.50}\)

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

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

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

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

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1\)

\(\Rightarrow B>48\)

a=8/9+15/16+24/25+....+2499/2500

a=(1-1/9)+(1-1/16)+(1-1/25)+....+(1-1/2500)

a=1-1/9+1-1/16+1-1/25+....+1-1/2500

a=(1+1+...+1)-(1/9+1/16+1/25+....+1/2500)

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

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

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

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

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

Bạn Phan Văn Hiếu ơi cho mình hỏi A là số nào vậy? Mà đề là chứng minh B<48 chứ

Bạn vào https://sites.google.com/site/toantieuhocpl/20-tinh-nhanh sẽ có đấy.

=2.4/3^2.3.5/4^2.4.6/5^2.....49.51/50^2

=(2.3.4.....49).(4.5.6.....51)/(3.4.5.....50).(3.4.5.....50)

=2.51/50.3

=17/25

giỏi hậy