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

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)

\(=\frac{1}{100}\)

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

Đặt : \(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}\)

\(A=\frac{1.2.3.4.....99}{2.3.4.5.....100}\)

\(A=\frac{1}{100}\)

Vậy : \(A=\frac{1}{100}\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}....\frac{99}{100}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(=\frac{1}{100}\)

\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{100}\left(1+2+3+....+100\right)\)

\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+\frac{1}{4}.\frac{4\left(4+1\right)}{2}+.....+\frac{1}{100}.\frac{100\left(100+1\right)}{2}\)

\(=1+\frac{2+1}{2}+\frac{3+1}{2}+....+\frac{100+1}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{101}{2}\)

\(=\frac{2+3+4+....+101}{2}\)

\(=\frac{\frac{101\left(101+1\right)}{2}-1}{2}=5150.5\)

\(S=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{100}\left(1+2+3+....+100\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+.....+\frac{1}{100}.\frac{100.101}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+.....+\frac{101}{2}\)

\(=\frac{2+3+4+....+101}{2}\)

\(=\frac{\frac{101.102}{2}-1}{2}\)

\(=2575\)

Vậy \(S=2575\)

Ta có:

 \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

nha

\(\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right)\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{1}{7}\right)...\left(1-1\frac{3}{7}\right)\left(1-1\right)\)

\(=\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)...\left(1-1\frac{3}{7}\right).0\)

\(=0\)

Trong dãy nhất định có \(\left[1-\frac{7}{7}\right]=0\)nên tích dãy trên là 0