\(E=\frac{2}{1\cdot3}\cdot\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
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TA CÓ:\(1\cdot3\cdot....\cdot99=\frac{\left(1\cdot3\cdot...\cdot99\right)\left(2\cdot4\cdot...\cdot100\right)}{2\cdot4....\cdot100}=\frac{1\cdot2\cdot3\cdot....\cdot100}{2\cdot2\cdot2\cdot...\cdot2\left(50\right)\cdot1\cdot2\cdot3\cdot..\cdot50}\)
\(=\frac{51\cdot52\cdot...\cdot100}{2\cdot2\cdot2\cdot...\cdot2}=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}\)(ĐPCM)
\(\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)-x\)\(=\frac{-100}{99}\)
\(\left(1-\frac{1}{99}\right)-x=\frac{-100}{99}\)
\(\frac{98}{99}-x=\frac{-100}{99}\)
\(x=\frac{98}{99}-\left(-\frac{100}{99}\right)\)
\(x=\frac{198}{99}=2\)
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ĐÚNG THÌ K CHO MK NHA!
\(2S=\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\)
\(2S=2-\frac{2}{99}\)
\(2S=\frac{196}{99}\)
\(S=\frac{196}{99}\cdot\frac{1}{2}=\frac{98}{99}\)
Ta có: S=2/1.3+2/3.5+...+2/97.99
S= 2/2.(1-1/3+1/3-1/5+...+1/97-1/99)
S= 1-1/99=98/99
\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3}{1}=8\)
Vậy A = 8
A = 2/3*5 + 2/5*7 + 2/7*9 + ... + 2/97*99
A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
A = 1/3 - 1/99
A = 32/99
\(A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}\)
\(A=\frac{32}{99}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(E=\frac{1}{1}-\frac{1}{99}\)
\(E=\frac{98}{99}\)
E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
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