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\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\) Chứng tỏ rằng A < 2
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1 /4 2 < 1/3.4
.. .........................
1/502 < 1/49.50
=> A < 1/12 + 1/1.2 + 1/2.3 + 1/3.4+......+1/49.50
=> A < 1 + (1-1/50)
=> A < 1+49/50
=> A < 99/55 <2
=> A < 2
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1 /4 2 < 1/3.4
.. .........................
1/502 < 1/49.50
=> A < 1/12 + 1/1.2 + 1/2.3 + 1/3.4+......+1/49.50
=> A < 1 + (1-1/50)
=> A < 1+49/50
=> A < 99/55 <2
=> A < 2
Ta có:
1/2 + 1/3 + 1/4 + ... + 1/15 + 1/16 = (1/2 + 1/3 + 1/4 + 1/5) + (1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11) + (1/12 + 1/13 + 1/14) + (1/15 + 1/16)
Vì 1/6 + 1/7 + 1/8 < 3x 1/6 = 1/2
1/9 + 1/10 + 1/11 <3x1/9 = 1/3
1/12 + 1/13 +1/14 < 3x1/12 = 1/4
1/15 + 1/16 < 3 x 1/15 = 1/5
Nên A < 2 x (1/2 + 1/3 + 1/4 + 1/5) < 2 x (1/2 + 1/2 + 1/4 + 1/4) =3 (1)
Lập luận tương tự có:
A = ( 1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12) + (1/13 + 1/14 + 1/15 + 1/16) > (1/2 + 1/3 + 1/4) + 4 x 1/8 + 4 x 1/ 12 + 4 x 1/16
Hay A > 2 x (1/2 + 1/3 + 1/4) > 2 x (1/2 + 1/4 + 1/4) = 2 (2)
Từ (1) và (2) ta có 2 < A < 3. Vậy A không phải là số tự nhiên.
Đặt \(Q=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{400}{401}\)
Áp dụng tính chất \(\frac{a}{b}< \frac{a+m}{b+m}\left(a,b,m\inℕ^∗\right)\)ta có
\(\frac{1}{2}< \frac{1+1}{2+1}=\frac{2}{3}\)
\(\frac{2}{3}< \frac{2+1}{3+1}=\frac{3}{4}\)
...
\(\frac{399}{400}< \frac{399+1}{400+1}=\frac{400}{401}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{399}{400}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{400}{401}\)
hay P < Q
=> \(P^2< P.Q\)
\(P^2< \frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{399}{400}.\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{400}{401}\)
\(P^2< \frac{1.2.3.4.....400}{2.3.4.5.....401}\)
\(P^2< \frac{1}{401}< \frac{1}{400}< \left(\frac{1}{20}\right)^2\)
Vì P và 1/20 có cùng dấu
\(\Rightarrow P< \frac{1}{20}\)
Bài 1:
Ta thấy:
\(\frac{1}{2}>\frac{1}{6};\frac{1}{3}>\frac{1}{6};\frac{1}{4}>\frac{1}{6};\frac{1}{5}>\frac{1}{6};\frac{1}{6}=\frac{1}{6}\)
\(=>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\)
\(=>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{5}{6}\)
Bài 2:
Đặt \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+...+\frac{1}{1517}\)
Ta thấy \(\frac{1}{5}=\frac{1}{1.5};\frac{1}{45}=\frac{1}{5.9};\frac{1}{117}=\frac{1}{9.13}\)
Theo quy luật như vậy ta có các số tiếp theo là:
\(\frac{1}{13.17}=\frac{1}{221};\frac{1}{17.21}=\frac{1}{357};\frac{1}{21.25}=\frac{1}{525};\frac{1}{25.29}=\frac{1}{725};...\)
Ta có \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+...+\frac{1}{1517}\)
\(=>A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{27.31}\)
\(=>4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{27.31}\)
\(=>4A=\frac{5-1}{1.5}+\frac{9-5}{5.9}+\frac{13-9}{9.13}+...+\frac{31-27}{27.31}\)
\(=>4A=\frac{5}{1.5}-\frac{1}{1.5}+\frac{9}{5.9}-\frac{5}{5.9}+\frac{13}{9.13}-\frac{9}{9.13}+...+\frac{31}{27.31}-\frac{27}{27.31}\)
\(=>4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{27}-\frac{1}{31}\)
\(=>4A=1-\frac{1}{31}=\frac{30}{31}=>A=\frac{30}{31}.\frac{1}{4}=\frac{15}{62}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(2A+A=\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\right)\)
\(3A=1-\frac{1}{64}\)
\(3A=\frac{63}{64}\Rightarrow A=\frac{63}{64}\div3=\frac{21}{64}< \frac{1}{3}\)
\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{101}}\)
\(2A-A=\frac{1}{2^{101}}-\frac{1}{2}\)
\(\Rightarrow A=\frac{1}{2^{101}}-\frac{1}{2}\)
\(\Rightarrow A>0\) ( đpcm )
Bài này phải làm như thế này nha lần trước tui làm nhầm sorry
Study well
\(\frac{1}{2\times2}+\frac{1}{3\times3}+....+\frac{1}{20\times20}