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

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

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

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: \(A=\frac{101}{200}\)

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right).....\left(1-\dfrac{1}{10000}\right)\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)

\(=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4....100.100}\)

\(=\dfrac{\left(1.2.3...99\right)}{2.3.4....100}\cdot\dfrac{3.4.5...101}{2.3.4....100}=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)

Ta có: \(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{10000}\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-9999}{10000}\)

\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\)

\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot1111\cdot9}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)

\(=\dfrac{101}{200}\)

la nho hon

`(1-1/4) \times (1-1/9) \times (1-1/16) \times ... \times (1-1/10000)`

`= 3/4 \times 8/9 \times 15/16 \times ... \times 9999/10000`

`= (3 \times 8 \times 15 \times ... \times 9999)/(4 \times 9 \times 16 \times ... \times 10000)`

`=`\(\dfrac{\left(1\times3\right)\times\left(2\times4\right)\times\left(3\times5\right)\times...\times\left(99\times101\right)}{\left(2\times2\right)\times\left(3\times3\right)\times\left(4\times4\right)\times...\times\left(100\times100\right)}\)

`=` \(\dfrac{\left(1\times2\times3\times...\times99\right)\times\left(3\times4\times5\times...\times101\right)}{\left(2\times3\times4\times...\times100\right)\times\left(2\times3\times4\times...\times100\right)}\)

`=` \(\dfrac{1\times101}{2\times100}\)

`= 101/200.`

(1-1/4) x ( 1-1/9) x ( 1- 1/16) x....x (1 - 1/10000)

= 3/4 x 8/9 x 15/16 x ... x 9999/10000

= (1x3/2x2) x ( 2x4/3x3 ) x ( 3 x 5 / 4x4 ) x ... x ( 99 x 101 / 100x 100 )

= \(\dfrac{\left(1.2.3...99\right).\left(3.4.5.101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}\)

= 1 x 101 / 100 x 2

= 101/200