\(1\cdot2+2\cdot3+3\cdot4+....+99\cdot100=\)
Những câu hỏi liên quan
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\times\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
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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\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
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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tính:\(\frac{1\cdot98+2\cdot97+3\cdot96+...+97\cdot2+98\cdot1}{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}\)
Tính tổng:
\(S=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{99\cdot100\cdot101}\)
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)
\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)
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\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+\dfrac{5}{3\cdot4}+.....+\dfrac{5}{98\cdot99}+\dfrac{5}{99\cdot100}\)
Giải:
\(\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+...+\dfrac{5}{98.99}+\dfrac{5}{99.100}\)
\(=5.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=5.\left(1-\dfrac{1}{100}\right)\)
\(=5.\dfrac{99}{100}\)
\(=\dfrac{99}{20}\)
Chúc em học tốt!
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Giải:
=5.(11.2+12.3+13.4+...+198.99+199.100)=5.(11.2+12.3+13.4+...+198.99+199.100)
=5.(1−1100)=5.(1−1100)
=9920=9920
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\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+................................................+\frac{3}{99\cdot100}\)
Đặt A = \(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{99.100}\)
\(\frac{1}{3}A\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{3}A\)\(=1-\frac{1}{100}\)
=> \(\frac{1}{3}A=\frac{99}{100}\)
=> A = \(\frac{99}{100}.3=\frac{297}{100}\)
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\(\frac{3}{1.2}+\frac{3}{2.3}+..................+\frac{3}{99.100}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+..................+\frac{1}{99.100}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{99}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)
\(=\frac{297}{100}\)
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\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+\frac{3}{3}-\frac{3}{4}...+\frac{3}{99}-\frac{3}{100}\)
\(=\frac{3}{1}-\frac{3}{100}\)
\(=\frac{297}{100}\)
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Tính
\(A=1\cdot2^2+2\cdot3^2+3\cdot4^2+.....+99\cdot100^2\)
\(1.2^2+2.3^2+...+99.100^2\)
\(=1.2\left(3-1\right)+2.3\left(4-1\right)+...+99.100\left(101-1\right)\)
\(=1.2.3-1.2+2.3.4-2.3+...+99.100.101-99.100\)
\(=\left(1.2.3+2.3.4+...+99.100.101\right)\)\(-\left(1.2+2.3+...+99.100\right)\)
Chúc học tốt
1)A=\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+.....+\dfrac{5}{99\cdot100}\)
C=\(1\cdot2\cdot3+2\cdot3\cdot4++3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7+6\cdot7\cdot8+7\cdot8\cdot9+8\cdot9\cdot10\)
D=\(1^2+2^2+3^2+...+99^2+100^2\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
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\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
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\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
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\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{4}{1}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
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