P=(1+\(\frac{1}{1\cdot3}\))(1+\(\frac{1}{2\cdot4}\))(1+\(\frac{1}{3\cdot5}\)).....(1+\(\frac{1}{99\cdot101}\)
Những câu hỏi liên quan
tính hợp lí :
B=\(\frac{1\cdot4}{2\cdot3}+\frac{2\cdot5}{3\cdot4}+\frac{3\cdot6}{4\cdot5}+.....+\frac{98\cdot101}{99\cdot100}\)
1) \(2x-\frac{4}{3}-\frac{4}{15}-\frac{4}{35}-\frac{4}{63}-\frac{4}{99}=\frac{15}{17}\)
2)\(\frac{10}{1\cdot2\cdot3}+\frac{10}{2\cdot3\cdot4}+\frac{10}{3\cdot4\cdot5}+.....+\frac{10}{100\cdot101\cdot102}\)
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
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Q=( \(1+\frac{1}{1\cdot3}\)) * ( \(1+\frac{1}{2\cdot4}\)) * ( \(1+\frac{1}{3\cdot5}\)) * ... * (\(1+\frac{1}{99\cdot101}\))
\(Q=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(Q=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{10000}{99.101}\)
\(Q=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{100.100}{99.101}\)
\(Q=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}\)
\(Q=100.\frac{2}{101}\)
\(Q=\frac{200}{101}\)
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\(Q=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)
\(Q=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{10000}{99.101}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{100.100}{99.101}\)
\(Q=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4.5...101}\)
\(Q=100.\frac{2}{101}\)
\(Q=\frac{200}{101}\)
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Cho S_1-S_2+S_3-S_4+S_5frac{m}{n} với m, n nguyên tố cùng nhau. Biết:S_1frac{1}{2}+frac{1}{3}+frac{1}{4}+frac{1}{5}+frac{1}{6}S_2frac{1}{2cdot3}+frac{1}{2cdot4}+frac{1}{2cdot5}+frac{1}{2cdot6}+frac{1}{3cdot4}+frac{1}{3cdot5}+frac{1}{3cdot6}+frac{1}{4cdot5}+frac{1}{4cdot6}+frac{1}{5cdot6}S_3frac{1}{2cdot3cdot4}+frac{1}{2cdot3cdot5}+frac{1}{2cdot3cdot6}+frac{1}{2cdot4cdot5}+frac{1}{2cdot4cdot6}+frac{1}{2cdot5cdot6}+frac{1}{3cdot4cdot5}+frac{1}{3cdot4cdot6}+frac{1}{3cdot5cdot6}+frac{1}{4cdot5cdot6}...
Đọc tiếp
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{99\cdot100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\\ =\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.........+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+......+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\\ =\frac{49}{100}\)
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\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{99}.\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
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\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\times\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Tìm giá trị của k.
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
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Tính
A=\(\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot....\cdot\left(1-\frac{1}{1326}\right)\)
B=\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\frac{1}{99\cdot101}\right)\)
Tính tổng A=\(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot5\cdot6}+...+\frac{1}{27\cdot28\cdot29\cdot30}\)
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
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\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
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Ta có:3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+.............+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1353}{8120}\Rightarrow A=\frac{451}{8120}\)
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Xem thêm câu trả lời
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Số k trong đẳng thức trên có giá trị là ?
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Rightarrow k=2\)
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