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Ta có : \(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-......-\frac{1}{2^{10}}\)
\(=\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-.....-\frac{1}{2^{10}}< \frac{1}{2}\) (đề sai)
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)(1)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}\)
Thay A vào (1)
\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)
\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Ta có: 210 < 211
\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2015^2}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2015}\)
\(\Rightarrow A< \approx0,75\)
Vậy.....
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt \(B=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: 1 = 1 ; \(\frac{1}{2^2}< \frac{1}{1.2}\); \(\frac{1}{3^2}< \frac{1}{2.3}\); .... ; \(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}< 2\)\(\Rightarrow B< 2\)
\(\Rightarrow A=\frac{1}{2^2}.B< \frac{1}{2^2}.2=\frac{1}{2}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
\(\Rightarrow\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)
Chứng minh rằng
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}\) < \(\frac{3}{4}\)
1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)
Nhận xét:
\(\frac{1}{2^2}<\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3^2}<\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}\)
....
\(\frac{1}{10^2}<\frac{1}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
Tính tổng ta có:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}<\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}<1\)
đặt A=1/1.2+1/2.3+...+1/9.10
B=1/2^2+1/3^2+...+1/10^2
ta có:B=1/2^2+1/3^2+...+1/10^2<A=1/1.2+1/2.3+...+1/9.10
mà A=1/1.2+1/2.3+...+1/9.10
=1-1/2+1/2-1/3+...+1/9-1/10
=1-1/10<1
=>A<B<1
=>A<1
C2:\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{10^2}\)<\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
=\(1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}<1\)
=\(\frac{9}{10}<1=\frac{9}{10}<\frac{10}{10}\)
Vậy \(\frac{9}{10}<1\)
C2:Ta có:\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}<\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
=\(1-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)<1
=9/10<1=9/10<10/10
Vậy 9/10<1
1/2^2+1/3^2+1/4^2+....+1/10^2<1
Đặt là A
A<B=1/1.2+1/2.3+1/3.4+...+1/9.10
B=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10
B=1-1/10=910<1
Do A< B<1
Suy ra A<1
TA có:1/2^2+1/3^2+1/4^2+...+1/10^2<1/1.2+1/2.3+1/3.4+...+1/9.10=1-1/10=9/10
Do 9/10<1 suy ra 1/2^2+1/3^2+...+1/10^2