B=(1-1/4).(1-1/9).(1-1/16)...(1-1/100^2)
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NT
0
9 tháng 8 2016
C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
TN
10 tháng 8 2016
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
NT
1
17 tháng 3 2023
\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)
NV
13 tháng 6 2018
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(B=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\)
\(B=\frac{3.8.15...9999}{2^2.3^2.4^2...100^2}\)
\(B=\frac{1.3.2.4.3.5...99.101}{2.2.3.3.4.4...100.100}\)
\(B=\frac{\left(1.2.3...99\right).\left(3.4.5...101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}\)
\(B=\frac{1.101}{100.2}\)
\(B=\frac{101}{200}\)
\(C=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right).\left(1+\frac{1}{100}\right)\)
\(C=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}.\frac{101}{100}\)
\(C=\frac{3.4.5...100.101}{2.3.4...99.100}\)
\(C=\frac{101}{2}\)
Dấu . là dâú x nha
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100^2}\right)\)
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{100^2}\)
\(B=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(B=\frac{1\cdot101}{100\cdot2}=\frac{101}{200}\)
S = 1 - 1/2² - 1/3² - 1/4² -.. - 1/100²
- - -
Có: 1/k² < 1/(k-1)k = 1/(k-1) - 1/k (với mọi k nguyên, k > 1)
1/2² < 1 /1.2 = 1/1 - 1/2
1/3² < 1 /2.3 = 1/2 - 1/3
...
1/10² < 1 /9.100 = 1/9 - 1/100
+ + + cộng vế lại + + +
1/2² + 1/3² +..+ 1/10² < 1 - 1/100
=> -1/2² - 1/3² - .. - 1/100² > -1 + 1/100
=> 1 - 1/2² - 1/3² - .. - 1/100² > 1/100 > 0 (đpcm)
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