\(\text{Cho }A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}\text{ Chứng tỏ }A< \frac{3}{4}\)
Những câu hỏi liên quan
chứng minh rằng
a)
\(\frac{1-2\text{s}in^2x}{2cot\left(\frac{\pi}{4}+\alpha\right).c\text{os}^2\left(\frac{\pi}{4}-\alpha\right)}=1\)
b)
\(\frac{\frac{\sqrt{3}}{2}c\text{os}2\text{a}-\frac{1}{2}sin2\text{a}}{1-\frac{1}{2}c\text{os}2\text{a}-\frac{\sqrt{3}}{2}sin2\text{a}}=tan\left(a+\frac{\pi}{4}\right)\)
chứng tỏ rằng: A= \(\frac{\text{1}}{2\text{1}}\) +\(\frac{2}{3\text{1}}\) +...+ \(\frac{20\text{1}3}{20\text{1}4}\) < 1
sửa lại đề : Chứng tỏ rằng : A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)
bài làm
A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}\)
A = \(\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2014-1}{2014!}\)
A = \(1-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2014}{2014!}-\frac{1}{2014!}\)
A = \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2013!}-\frac{1}{2014!}\)
A = \(1-\frac{1}{2014!}< 1\)
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Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}\) . Chứng tỏ \(A< \frac{3}{4}\)
\(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
.......................................
\(\frac{1}{2014^2}=\frac{1}{2014\cdot2014}< \frac{1}{2013\cdot2014}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2013\cdot2014}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow A< 1-\frac{1}{2014}=\frac{2013}{2014}\)
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Trần Nhật Dương Chứng minh \(A< \frac{3}{4}\) mà :))
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Ta có: \(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
........................
\(\frac{1}{2014^2}=\frac{1}{2014.2014}< \frac{1}{2013.2014}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{3}{4}-\frac{1}{2014}< \frac{3}{4}\)
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Tính :
a) \(\text{A}=\left(1\times2\right)^{-1}+\left(2\times3\right)^{-1}+...+\left(2014\times2015\right)^{-1}\).
b) \(\text{B}=\frac{2018+\frac{2017}{2}+\frac{2016}{3}+\frac{2015}{4}+...+\frac{2}{2017}+\frac{1}{2018}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2018}+\frac{1}{2019}}\).
a, Tính nhanh :
\(\frac{2009\times(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008})}{2008-\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{2006}{2007}+\frac{2007}{2008}\right)}\)
b, Cho \(\text{Q}=2+2^2+2^3+...+2^{10}\). Chứng tỏ rằng \(Q⋮3\).
A=\(\frac{1}{2^2}\) +\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\)+...+\(\frac{1}{2014^2}\). Chứng tỏ A < \(\frac{3}{4}\)
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)
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Cho A =\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{2014^2}\).Chứng tỏ A < \(\frac{3}{4}\)
Ta có:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2014^2}\)
\(< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2013\cdot2014}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{2013}-\frac{1}{2014}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)
\(=\frac{3}{4}-\frac{1}{2014}\)
\(< \frac{3}{4}\)
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Cho A=\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...........+\frac{1}{1+2+3+\text{...+99}}\)
B=\(1+\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{49^2}\)
Chứng minh A<B
Ta có :
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+3}+...+\frac{1}{1+2+3+...+99}\)
\(A=\frac{1}{\frac{2\left(2+1\right)}{2}}+\frac{1}{\frac{3\left(3+1\right)}{2}}+\frac{1}{\frac{4\left(4+1\right)}{2}}+...+\frac{1}{\frac{99\left(99+1\right)}{2}}\)
\(A=\frac{2}{2\left(2+1\right)}+\frac{2}{3\left(3+1\right)}+\frac{2}{4\left(4+1\right)}+...+\frac{2}{99\left(99+1\right)}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(A=2.\frac{49}{100}\)
\(A=\frac{49}{50}\)
Lại có :
\(\frac{1}{2^2}>\frac{1}{2.3}\)
\(\frac{1}{3^2}>\frac{1}{3.4}\)
\(\frac{1}{4^2}>\frac{1}{4.5}\)
\(............\)
\(\frac{1}{49^2}>\frac{1}{49.50}\)
\(\Rightarrow\)\(B=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{49^2}>1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(B>1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(B>1+\frac{1}{2}-\frac{1}{50}\)
\(B>1+\frac{12}{25}=\frac{37}{25}=\frac{74}{50}>\frac{49}{50}=A\)
\(\Rightarrow\)\(B>A\)
Vậy \(A< B\)
Chúc bạn học tốt ~
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Chứng minh rằng :
a, \(\text{A}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{2}{4^{\text{2}}}+...+\frac{1}{(n-1)^2}+\frac{1}{n^2}< 1\) ;
b, \(\text{B}=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\).
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
........
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
Đpcm
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b)B=1/4(1/2^2+1/3^2+...+1/n^2)=1/4*A<1/4
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tìm 2 số 3:
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