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1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
\(\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{2^2}\right)+...+\left(1+\dfrac{1}{2^{50}}\right)\)
= \(\left(1+1+1+...+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)(50 số 1 )
= \(50+\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\right)\)
A =\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{50}}\)
⇒ 2A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\)
⇒ 2A - A =\(1-\dfrac{1}{2^{50}}\)
=50+1-\(\dfrac{1}{2^{50}}\)=51-\(\dfrac{1}{2^{50}}>3\)
Ta có :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{2}{5}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.................+\dfrac{99}{100}}\)
\(=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+.............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+............+1-\dfrac{1}{100}}\)
\(=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+...........+\dfrac{2}{100}\right)}{\left(1+1+.........+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{100}\right)}\)
\(=\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..........+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.........+\dfrac{1}{100}\right)}\)
\(=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+..........+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}=2\rightarrowđpcm\)
a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)
\(=\dfrac{3.4.5...100}{2.3.4...99}\)
\(=\dfrac{100}{2}=50\)
a,
\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)
b,
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)
c,
\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)
Lời giải:\(A=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-4\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)
\(A=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
\(\Rightarrow A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right)}{2.3.4....99.100}\)
\(\Rightarrow A=\dfrac{1}{100}\)
A = - 522 - { - 222 - [ - 122 - (100 - 522) + 2022] }
A = - 522 - { -222 - [- 122 - 100 + 522 ] + 2022}
A = - 522 - { -222 - { - 222 + 522 } + 2022}
A = - 522 - {- 222 + 222 - 522 + 2022}
A = -522 + 522 - 2022
A = - 2022
B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)
B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2
B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2
B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)
B = \(\dfrac{2+3+4+...+21}{2}\)
B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)
B = \(\dfrac{23\times20:2}{2}\)
B = \(\dfrac{23\times10}{2}\)
B = 23
1) Ta có
\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)
\(C=\dfrac{1}{2022}\)
2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)