A = 3/4 + 8/9 + 15/16 + ... + 399/400
A = 1 - 1/4 + 1 - 1/9 + 1 - 1/16 + ... + 1 - 1/400
A = (1 + 1 + 1 + ... +1) - (1/4 + 1/9 + 1/16 + ... + 1/400)
A = 19 - (1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20)
đặt b = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/20.20
có 1/2.2 < 1/1.2 ; 1/3.3 < 1/2.3 ; ... 1/20.20 < 1/19.20
=> b < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/19.20
=> b < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/19 - 1/20
=> b < 1 - 1/20
=> b < 1
mà A = 19 - b
=> A > 18
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+.....+\frac{20^2-1}{20^2}\)
\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{20^2}\right)\)
\(>19-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{19\cdot20}\right)\)
\(=19-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{19}-\frac{1}{20}\right)\)
\(=19-\left(1-\frac{1}{20}\right)\)
\(>19-1=18\)
\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+...+\frac{20^2-1}{20^2}\)
\(A=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+...+\frac{20^2}{20^2}-\frac{1}{20^2}\)
\(A=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)
\(B< 1-\frac{1}{20}< 1\)
\(B< 1\)
\(\Rightarrow A>19-1\)
\(\Rightarrow A>18\)
Ai kick sai cho mik cho bt mik sai ở đâu zợ ?
Ta có :
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{399}{400}\)
\(=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+...+\frac{399}{20^2}\)
\(A=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{20^2}\right)\)
\(=1+1...+1-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)
\(=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\)> \(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)=19-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=19-\left(1-\frac{1}{20}\right)=18+\frac{1}{20}>18\)
\(\Rightarrow A>\frac{1}{18}\)
\(A=\frac{3}{4}+\frac{8}{9}+...+\frac{399}{400}\)
\(\Rightarrow A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+...\left(1-\frac{1}{400}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{20^2}\right)\)
\(\Rightarrow A=19-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\right)\) \(\left(1\right)\)
Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{20^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{20^2}< \frac{1}{19.20}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)
\(\Rightarrow B< 1-\frac{1}{20}< 1\)
\(\Rightarrow B< 1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow A>19-1=18\) \(\Rightarrow A>18\)
