Cho A =\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{99^2}\)
CMR A<1
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ta có \(\frac{1}{1^2}<\frac{1}{1.2},\frac{1}{2^2}<\frac{1}{2.3},.........,\frac{1}{100^2}<\frac{1}{100.101}\)
=> A <\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{100.101}\)
dến đây bạn tự tính nha mình tính đc bằng
A < \(\frac{1}{1}-\frac{1}{101}\)
bây giờ tự lập luận là đc , đơn giản mà
kết bạn vs mình cũng đc , có bài nào thì mình bày cho
\(VP=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(VP=\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)
\(VP=\frac{2}{2}-\frac{1}{2}+\frac{3}{3}-\frac{1}{3}+\frac{4}{4}-\frac{1}{4}+...+\frac{100}{100}-\frac{1}{100}\)
\(VP=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)
\(VP=100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=VT\) ( đpcm )
Mk nghĩ \(VT=100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\) bn xem lại đề có nhầm ko
Chúc bạn học tốt ~
\(A=\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+...........+\frac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{99}}\)
\(\Leftrightarrow2^{99}.A=2^{99}-1\left(đpcm\right)\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\)\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\)=\(\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\)\(|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{1+\frac{1}{2^2}+\frac{1}{\left(-3\right)^2}}\)\(=|\frac{1}{1}+\frac{1}{2}+\frac{1}{-3}|=1+\frac{1}{2}-\frac{1}{3}\)
Tương tự ta có M=\(1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{99}-\frac{1}{100}\)=\(98+\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)
a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)
=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
=\(1-\frac{1}{100!}< 1\)
\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)
=\(2-\frac{1}{99}-\frac{1}{100}< 2\)
\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(A< 1-\frac{1}{99}=\frac{98}{99}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
....
\(\frac{1}{99^2}< \frac{1}{98.99}=\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\)\(\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow A< 1-\frac{1}{99}\)
\(\Rightarrow A< 1\)
Tham khảo nha !!!