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\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
a)đặt B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)
từ (1),(2),(3) =>A<2
b,c tự làm
Ta có: \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};....;\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}< 1\left(đpcm\right)\)
Ta có : \(\frac{1}{2^2}=\frac{1}{4}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{9}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{16}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}=\frac{1}{10000}< \frac{1}{99.100}\)
Suy ra : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy ta có đpcm
Ta có:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2014^2}\)
\(< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2013\cdot2014}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{2013}-\frac{1}{2014}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)
\(=\frac{3}{4}-\frac{1}{2014}\)
\(< \frac{3}{4}\)
Sorry bạn nha , mình bấm nhầm nút
\(A=\frac{5}{4}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A< \frac{5}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< \frac{5}{4}+\frac{1}{2}-\frac{1}{100}< \frac{5}{4}+\frac{1}{2}=\frac{7}{4}\)
\(\Rightarrow\)\(A< \frac{7}{4}\)
Vậy , \(\frac{5}{4}< A< \frac{7}{4}\left(ĐPCM\right)\)
BÀI KHÓ CỦA TRƯỜNG MÌNH ĐÓ THI HK2
GIÚP MÌNH NHÉ!!!!!!THANKS!!!!!!
Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
\(=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\right)\)
Nhận xét: \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(\frac{1}{5^2}< \frac{1}{4\cdot5}\)
...
\(\frac{1}{2014^2}< \frac{1}{2013\cdot2014}\)
Do đó: \(\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\right)< \frac{1}{4}+\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2013\cdot2014}\right)\)
\(\Leftrightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(\Leftrightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{3019}{4028}\)
mà \(\frac{3019}{4028}< \frac{3021}{4028}=\frac{3}{4}\)
nên \(A< \frac{3}{4}\)(đpcm)
Đặt A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)< A=\(\frac{1}{1.2}+\frac{1}{2.3}+........+\frac{1}{99.100}\)
A=\(1-\frac{1}{100}=\frac{100}{100}-\frac{99}{100}\)=\(\frac{1}{100}<\frac{3}{4}=\frac{4}{400}<\frac{300}{400}\)
Vậy A<3/4