Gọi \(A=1×2+2×3+..+99×100\)
\(3A=1.2.3+2.3.3+...+999.100.3=1.2\left(3-0\right)+2.3\left(4-1\right)+...+98.99\left(100-97\right)=1.2.3+2.3.4-1.2.3+...-98.99.100-99.100.101=99.100.101\)
\(A=\frac{99.100.101}{3}=333300\)
Gọi \(A=1×2+2×3+..+99×100\)
\(3A=1.2.3+2.3.3+...+999.100.3=1.2\left(3-0\right)+2.3\left(4-1\right)+...+98.99\left(100-97\right)=1.2.3+2.3.4-1.2.3+...-98.99.100-99.100.101=99.100.101\)
\(A=\frac{99.100.101}{3}=333300\)
\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+\dfrac{5}{3\cdot4}+.....+\dfrac{5}{98\cdot99}+\dfrac{5}{99\cdot100}\)
\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+................................................+\frac{3}{99\cdot100}\)
Tính :
\(A=1\cdot2+2\cdot3+.......+99\cdot100\)
\(choA=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100};B=\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+...+\frac{1}{52\cdot99}+\frac{1}{100\cdot51}\)
Tính nhanh : \(4-\frac{2}{1\cdot2}-\frac{2}{2\cdot3}-\frac{3}{3\cdot4}-.........-\frac{2}{99\cdot1000}\)
1) \(2x-\frac{4}{3}-\frac{4}{15}-\frac{4}{35}-\frac{4}{63}-\frac{4}{99}=\frac{15}{17}\)
2)\(\frac{10}{1\cdot2\cdot3}+\frac{10}{2\cdot3\cdot4}+\frac{10}{3\cdot4\cdot5}+.....+\frac{10}{100\cdot101\cdot102}\)
1) Tính
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
b)\(\left(\frac{15}{1\cdot2\cdot3}+\frac{15}{2\cdot3\cdot4}+\frac{15}{3\cdot4\cdot5}+.....+\frac{15}{18\cdot19\cdot20}\right)\cdot x=1\)
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
1.
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot1018}\)
b) \(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2017\cdot2018}+\frac{2}{2018\cdot2019}\)
c) \(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+......+\frac{4}{1999\cdot2000}\)
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