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\(S=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right)...\left(\frac{1}{81}-1\right).\left(\frac{1}{100}-1\right)\)
\(S=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}........\frac{-80}{81}.\frac{-99}{100}\)
\(-S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{80}{81}.\frac{99}{100}\)
\(-S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}........\frac{8.10}{9.9}.\frac{9.11}{10.10}\)
\(-S=\frac{1.3.2.4.3.5........8.10.9.11}{2.2.3.3.4.4.......9.9.10.10}\)
\(-S=\frac{\left(1.2.3......8.9\right).\left(3.4.5.......10.11\right)}{\left(2.3.4.......9.10\right).\left(2.3.4........9.10\right)}\)\(=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}=>S=\frac{-11}{20}\)
D= [(1-1/2)(1-1/3)...(1-1/25)]:[(1+1/2)(1+1/3)...(1+1/25)]
D= [1/2. 2/3. ... . 24/25]: [3/2. 4/3. ... . 26/25]
D= 1/25 : 2/26
D= 1/25 . 26/2= 13/25
Vậy D= 13/25
\(D=\left[\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{25}\right)\right]\)\(:\left[\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{25}\right)\right]\)
\(D=\left[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{24}{25}\right]:\left[\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{26}{25}\right]\)
\(D=\frac{1.2.3...24}{2.3.4...25}:\frac{3.4.5...26}{2.3.4...25}\)
\(D=\frac{1}{25}:13\)
\(D=\frac{1}{325}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)......\left(1-\frac{1}{81}\right)\left(1-\frac{1}{100}\right)\)
= \(-\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{80}{81}.\frac{99}{100}\)
=\(-\frac{1.3.2.4.3.5..............8.10.9.11}{2^2.3^2.4^2.......10^2}=-\frac{\left(1.2.3.....9\right)\left(3.4.5....11\right)}{2.3.4....10.2.3.4.....10}=-\frac{11}{20}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{120}{121}=\frac{3.8.15...120}{4.9.16...121}\)
\(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(11.11\right)}\)
\(=\frac{\left(1.2.3...10\right).\left(3.4.5...12\right)}{\left(2.3.4...11\right).\left(2.3.4...11\right)}=\frac{1.12}{11.2}=\frac{6}{11}\)
ta có :
A=\(\left(-\frac{3}{4}\right)\left(-\frac{8}{9}\right)\left(-\frac{15}{16}\right)...\left(-\frac{120}{121}\right)\)(có 10 số hạng)
= \(\frac{3\cdot8\cdot15\cdot...\cdot120}{4\cdot9\cdot16\cdot...\cdot121}=\frac{\left(1.3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(10\cdot12\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot11^2}=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot..\cdot11\right)\left(2\cdot3\cdot4\cdot..\cdot11\right)}\)
=\(\frac{12}{11\cdot2}=\frac{12}{22}\)
=\(-\frac{11}{20}\)