Question

[Hoang Anh Dũng]

(\(\frac{1}{2^2}-1\)).(\(\frac{1}{3^2}-1\)).(\(\frac{1}{4^2}-1\)).......(\(\frac{1}{100^2}-1\))

Answer 1

[★๖ۣۜßảo๖ۣۜPɦα♏๖ۣۜ[EηgĻïšħ☯€lub]★]

Ta có:\(-\left(n-1\right)\left(n+1\right)=-\left(n^2+n-n-1\right)=-n^2+1=1-n^2\)

Xét:\(A=...\)

\(=\frac{1-2^2}{2^2}\cdot\frac{1-3^2}{3^2}\cdot...\cdot\frac{1-100^2}{100^2}\)

Mà A có:99 số hạng => số lẻ,nên:

\(A=-\left(\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot...\cdot\frac{100^2-1}{100^2}\right)\)

\(=-\left(\frac{1\cdot3}{2^2}\cdot\frac{2\cdot4}{3^2}\cdot...\cdot\frac{99\cdot101}{100^2}\right)\)

\(=-\left(\frac{1}{2}\cdot\frac{101}{100}\right)=-\frac{101}{200}\)

Answer 2

[Tôi không biết]

A=\(\left(\frac{1}{4}-1\right)\)\(\left(\frac{1}{9}-1\right)\)\(\left(\frac{1}{16}-1\right)\)...\(\left(\frac{1}{10000}-1\right)\)

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

vì có 50 thừa số nên A= \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\)

=>A=\(\frac{3.8.15.....9999}{4.9.16.....10000}\)

A=\(\frac{\left(1.3\right)\left(2.4\right)\left(3.5\right).....\left(99.101\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right).....\left(100.100\right)}\)

A=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)

A=\(\frac{1.101}{100.2}\)

A=\(\frac{101}{200}\)