Cho A = 1/2^2 + 1/3^2 + 1/4^2 + ......... + 1/2014^2. Chứng tỏ A<3/4
Lưu ý: Dấu ^ là dấu mũ nhaaa:3 Các cậu giải giúp tớ vớiiii:4
cho A1-1/2^2-1/3^2-1/4^2-1/5^2...-1/2010^2. chứng tỏ A>1/2014
\(\text{Cho }A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}\text{ Chứng tỏ }A< \frac{3}{4}\)
\(n^2>\left(n-1\right)\left(n+1\right)\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).\)
Do đó: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2012.2014}+\frac{1}{2013.2015}=\)
\(=\frac{1}{2}[1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2015}]=\)
\(=\frac{1}{2}[1+\frac{1}{2}-\frac{1}{2014}-\frac{1}{2015}]=\frac{1}{2}[\frac{3}{2}-\frac{1}{2014}-\frac{1}{2015}]=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{2014}+\frac{1}{2015}\right)< \frac{3}{4}.\)
Bài 1:So sánh 20142014 + 1/20142015 + 1 và 20142013 + 1/20142014 + 1. Bài 2: a) chứng tỏ rằng: D=1/22 + 1/32 + 1/42 +....+1/102 < 1. b)chứng tỏ rằng: E=1/101+1/102+...+1/299+1/300>2/3.C)chứng tỏ rằng: F=1/5+1/6+1/7+...+1/17 < 2
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}\Leftrightarrow D< 1\left(đpcm\right)\)
Cho: A=1/2+(1/2)2+(1/2)3 +(1/2)4+......+(1/2)2013+(1/2)2014. Chứng tỏ A < 1
nhân A với 2:
Lấy A.2-A và ra A=1-(1/2)^2014<1
A = 1/4 + 1/9 + 1/10+ ... + 1/2014 mũ 2 hãy chứng tỏ A < 3/4
Lời giải:
$A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}$
$< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}$
$=\frac{1}{4}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2014-2013}{2013.2014}$
$=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}$
$=\frac{1}{4}+\frac{1}{2}-\frac{1}{2014}$
$< \frac{1}{4}+\frac{1}{2}=\frac{3}{4}$
Ta có đpcm.
Lần sau bạn lưu ý gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo)
Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}\) . Chứng tỏ \(A< \frac{3}{4}\)
\(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
.......................................
\(\frac{1}{2014^2}=\frac{1}{2014\cdot2014}< \frac{1}{2013\cdot2014}\)
\(\Rightarrow A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2013\cdot2014}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Rightarrow A< 1-\frac{1}{2014}=\frac{2013}{2014}\)
Trần Nhật Dương Chứng minh \(A< \frac{3}{4}\) mà :))
Ta có: \(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
........................
\(\frac{1}{2014^2}=\frac{1}{2014.2014}< \frac{1}{2013.2014}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2014^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{1}{2^2}+\frac{1}{2}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{2014}\)
\(\Leftrightarrow A< \frac{3}{4}-\frac{1}{2014}< \frac{3}{4}\)
Cho A = 1/5 + 1/5^2 + 1/5^3 +....+ 1/5^2014 . Chứng tỏ rằng A < 1/4
\(A=\frac{1}{5}+\frac{1}{5^2}+........+\frac{1}{5^{2014}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...........+\frac{1}{5^{2013}}\)
\(\Rightarrow5A-A=1+...........+\frac{1}{5^{2013}}-\frac{1}{5}+...........+\frac{1}{5^{2014}}\)
\(\Rightarrow4A=1-\frac{1}{5^{2014}}\)
\(\Rightarrow4A< 1\Rightarrow A< \frac{1}{4}\)
=> 5A = 1 + 1/5 +...+1/5^2013
=>4A= 1- 1/5^2014
=> 4A< 1 => A < 1/4
Cho A = (1+1/2+1/3+1/4+...+1/97+1/98). 20142015. Chứng tỏ A chia hết cho 11
cho A=(2014+1)(2014+2)....(2014 +2014) chứng tỏ rằng A chia hết cho 2^2014