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\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....\left(\frac{1}{101}-1\right)\)
\(=\frac{3}{4}.\frac{8}{9}.....\frac{120}{121}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{10.12}{11.11}\)
\(=\frac{\left(1.3\right).\left(2.4\right).....\left(10.12\right)}{\left(2.2\right).\left(3.3\right).....\left(11.11\right)}\)
\(=\frac{\left(1.2.3.....10\right).\left(3.4.5.....12\right)}{\left(2.3.4.....11\right).\left(2.3.4.....11\right)}\)
\(=\frac{1.12}{11.2}=\frac{6}{11}\)
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Ta có \(63,1.2-21,3.6=0,9.7.10.1,2-21.3,6\)
\(=6,3.1,2-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.1,2-6,3.1,2\)
\(=0\)
\(\Rightarrow\dfrac{\left(1+2+......+100\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+.....+99-100}=\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)0}{1-2+3-4+......+99-100}=0\)
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\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100^2}\right)\)
\(B=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{9999}{100^2}\)
\(B=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}\)
\(B=\frac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}\)
\(B=\frac{1\cdot101}{100\cdot2}=\frac{101}{200}\)
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)
♥Tomato♥
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\(=\frac{1}{1.3}.\frac{1}{2.4}...\frac{1}{9.11}=\frac{1}{1.2.3^2...9^2.10.11}\)
\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).....\left(\frac{1}{100}-1\right)\)
\(=\frac{3}{4}.\frac{8}{9}.....\frac{99}{100}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.....\frac{9.11}{10.10}\)
\(=\frac{\left(1.3\right).\left(2.4\right).....\left(9.11\right)}{\left(2.2\right).\left(3.3\right).....\left(10.10\right)}\)
\(=\frac{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}\)
\(=\frac{1.11}{10.2}=\frac{11}{20}\)