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\(A=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}....\frac{100^2-1}{100^2}\)
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}....\frac{99.101}{100^2}\)
\(A=\frac{1.3.2.4...99.100}{2.2.3.3...100.100}\)
\(A=\frac{1.2...99}{2.3....100}.\frac{3.4...101}{2.3...100}\)
\(A=\frac{1}{100}.\frac{101}{2}\)
\(A=\frac{101}{200}\)
\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{2013^2-1}{2013^2}.\frac{2014^2-1}{2014^2}\)
\(A=\frac{1.3.2.4.3.5....2012.2014.2013.2015}{2^2.3^2.4^2...2013^2.2014^2}\)
\(A=\frac{\left(1.2.3...2012.2013\right).\left(3.4.5...2014.2015\right)}{\left(2.3.4...2013.2014\right).\left(2.3.4...2013.2014\right)}\)(nhóm từng số ở trước và sau vào 2 nhóm khác nhau)
\(A=\frac{3.2015}{2014.2}\)
\(A=\frac{6045}{4028}\)
\(A=\frac{6045}{4028}\),nha bạn ,chúc bạn hok tốt ,love bạn nhìu ,cách làm giống như Monozono Nanami nha
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
\(P=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(\Rightarrow P=\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{9}{9}-\frac{1}{9}\right)\left(\frac{16}{16}-\frac{1}{16}\right)...\left(\frac{2500}{2500}-\frac{1}{2500}\right)\)
\(\Rightarrow P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)
\(\Rightarrow P=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
\(\Rightarrow P=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{\left(2.3.4...50\right)\left(2.3.4...50\right)}\)
\(\Rightarrow P=\frac{51}{50.2}=\frac{51}{100}>\frac{50}{100}=\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Ta có:
\(P=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{50^2}\right)\)
\(P=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{2500}\right)\)
\(P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{2499}{2500}\)
\(P=\frac{3.8.15.....2499}{4.9.16.....2500}\)
Tới chỗ này rồi tiếp tục rút gọn
Kết quả cuối cùng là: \(P>\frac{1}{2}\)
Xin lỗi nha, tớ ko có giỏi ở phần rút gọn.
\(P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)
\(P=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
\(P=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{\left(2.3.4...50\right)\left(2.3.4...50\right)}\)
\(P=\frac{1.51}{50.2}\)
\(P=\frac{51}{100}>\frac{1}{2}\)
Kết luận: \(P>\frac{1}{2}\)
Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(=\left(-\frac{1.3}{2.2}\right).\left(-\frac{2.4}{3.3}\right)...\left(-\frac{99.101}{100.100}\right)\)
\(=-\frac{1}{2}.\frac{101}{100}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
Vậy \(A< -\frac{1}{2}\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(=\frac{-3}{4}\cdot\frac{-8}{9}\cdot\frac{-15}{16}\cdot...\cdot\frac{-9999}{10000}\)
\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot...\cdot\frac{-99\cdot111}{100.100}\)
\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{99\cdot111}{100\cdot100}\)
\(=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot6\cdot...\cdot111\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot100\right)^2}\)
\(=\frac{101}{2\cdot100}\)
\(=\frac{101}{200}>\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{4}{2^2}\right)\left(\frac{1}{3^2}-\frac{9}{3^2}\right)...\left(\frac{1}{100^2}-\frac{10000}{100^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(\Rightarrow A=-\frac{3}{2^2}.\frac{8}{3^2}...\frac{9999}{100^2}\)
\(\Rightarrow A=-\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(\Rightarrow A=-\frac{\left(1.2...99\right)\left(3.4...101\right)}{\left(2.3...100\right)\left(2.3...100\right)}\)
\(\Rightarrow A=-\frac{101}{100.2}=\frac{-101}{200}< \frac{-100}{200}=\frac{-1}{2}\)
Vậy \(A< \frac{-1}{2}\)
\(A=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{100^2}\right)\)
\(A=-\left(\frac{1.3}{2.2}\right).\left(\frac{2.4}{3.3}\right)....\left(\frac{99.101}{100.100}\right)\)
\(A=-\left(\frac{1.2....99}{2.3...100}\right).\left(\frac{3.4....101}{2.3....100}\right)\)
\(A=-\left(\frac{1}{100}\right).\left(\frac{101}{2}\right)\)
\(A=\frac{-101}{200}>\frac{-1}{2}\)
Ta có: Q=(1-1/2^2).(1-1/3^2).....(1-1/40^2)
Q=3/2^2.8/3^2....1599/40^2
Q=(3/2.2).(8/3.3)...(1599/40.40)
Q=(1.3/2.2).(2.4/3.3)...(39.41/40.40)
Q=(1.2...39/2.3...40).(3.4...41/2.3...40)
Q=1/40.41/2
Q=41/80
Mà 41/80>40/80=1/2
=>Q > 1/2
\(Q=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{40^2}\right)\)
\(\Rightarrow Q=\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{9}{9}-\frac{1}{9}\right)...\left(\frac{1600}{1600}-\frac{1}{1600}\right)\)
\(\Rightarrow Q=\frac{3}{4}.\frac{8}{9}...\frac{1599}{1600}\)
\(\Rightarrow Q=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{39.41}{40.40}\)
\(\Rightarrow Q=\frac{\left(1.2.3...39\right)\left(3.4.5...41\right)}{\left(2.3.4...40\right)\left(2.3.4...40\right)}\)
\(\Rightarrow Q=\frac{41}{40.2}=\frac{41}{80}>\frac{40}{80}=\frac{1}{2}\)
Vậy \(Q>\frac{1}{2}\)