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A=1/(2x2)+1/(3x3)+...+1/(100x100)
Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1)
=> A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4
A=1/(2x2)+1/(3x3)+...+1/(100x100) Nhận thấy rằng n x n -1=n x n -n+n-1=n x (n-1)+n-1=(n-1) x (n+1) => A < 1/(2x2-1)+1/(3x3-1)+...+1/(100x100-1)=1/(1x3)+1/(3x5)+...+1/(99x101)=1/2-1/202<1/2<3/4
đáp số cuối cùng là 0 vì 1-1/4 = 0/4 = 0 x cho những số nào khác cũng bằng 0 thôi
\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times...\times\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{999}{10000}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times...\times\frac{99\times101}{100\times100}\)
\(=\frac{1}{2}\times\frac{101}{100}\)
\(=\frac{101}{200}\)
M = \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}.\right)\left(1-\frac{1}{16}\right)....\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}=\frac{3.8.15...9999}{4.9.16....10000}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)....\left(99.101\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)....\left(100.100\right)}\)
\(=\frac{\left(1.2.3...99\right).\left(3.4.5..101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)
\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
S=1/4+1/9+1/16+...+1/10000 = 1/2x 2 + 1/3x3+...+1/100x100 < 1/1x2 + 1/2x3 +...+ 1/9x10 = 1 - 1/2 + 1/2 - 1/3 +...+ 1/9 - 1/10 = 1- 1/10 < 1
Ta có: \(1-\frac{1}{4}=\frac{3}{4}=\frac{1}{2}.\frac{3}{2}\); \(1-\frac{1}{9}=\frac{8}{9}=\frac{2}{3}.\frac{4}{3}\); \(1-\frac{1}{16}=\frac{15}{16}=\frac{3}{4}.\frac{5}{4}\);
...; \(1-\frac{1}{10000}=\frac{9999}{10000}=\frac{99}{100}.\frac{101}{100}\)
=> \(A=\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}.\frac{3}{4}.\frac{5}{4}....\frac{99}{100}.\frac{101}{100}\). Nhận thấy; Tích của 2 số liền kề thì bằng 1
=> \(A=\frac{1}{2}.\frac{101}{100}=\frac{101}{200}\)
Đáp số: \(A=\frac{101}{200}\)