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Ta co \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^{10}}\)
=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
=\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
=\(\frac{1}{1^2}-\frac{1}{10^2}\)
=\(\frac{99}{100}\) < 1
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900
Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi
\(A=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot...\cdot1-\frac{2}{99\cdot100}\)
\(2A=1-\left(\frac{1}{2\cdot3}\cdot\frac{1}{3\cdot4}\cdot\frac{1}{4\cdot5}\cdot...\cdot\frac{1}{99\cdot100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}-\frac{1}{5}\cdot...\cdot\frac{1}{99}\cdot\frac{1}{100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(2A=1-\frac{49}{100}\)
\(2A=\frac{51}{100}\)
\(A=\frac{51}{100}:2\)
\(A=\frac{51}{200}\)
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5..100}\)
\(=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
\(=2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)...2\left(\frac{1}{2}-\frac{2}{99.100}\right)\)
\(=2^{89}.\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)\)
= \(\frac{2^{98}.4851}{100}\)
Ta có :
\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{19}{\left(9.10\right)^2}\)
\(=\)\(\frac{3}{1.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\)\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\)\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=\)\(1-\frac{1}{100}\)
\(=\)\(\frac{100}{100}-\frac{1}{100}\)
\(=\)\(\frac{100-1}{100}\)
\(=\)\(\frac{99}{100}\)
Vậy ...
Đặt A=\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+..........+\frac{19}{\left(9.10\right)^2}\)
A=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+.........+\frac{19}{9^2.10^2}\)
A=\(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...........+\frac{19}{81.100}\)
A=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-...............+\frac{1}{81}-\frac{1}{100}\)
A=\(\frac{1}{1}-\frac{1}{100}\)
A=\(\frac{99}{100}\)
Vậy tổng của \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+..........+\frac{19}{\left(9.10\right)^2}\)là \(\frac{99}{100}\)
Chúc bn học tốt