Cho A= \(\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2017}\)
Chứng minh A < \(\frac{3}{4}\)
Cho A= \(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2017}\)
Chứng minh A<\(\frac{3}{4}\)
A=1/(1+3)+1/(1+3+5)+1/(1+3+5+7)+...+1/(1+3+5+7+...+2017)
A=1/2^2+1/3^2+1/4^2+...+1/1009^2
2A=2/2^2+2/3^2+2/4^2+...+2/1009^2
Ta co :(x-1)(x+1)=(x-1)x+x-1=x^2-x+x-1=x^2-1<x^2
suy ra 2A<2/(1*3)+2/(3*5)+2/(5*7)+...+2/(1008*1010)
suy ra 2A <1-1/3+1/3-1/5+1/5-1/7+...+1/1008-1/1010
suy ra 2A<1-1/1010
suy ra 2A<2009/2010<1<3/2
suy ra 2A <3/2
suy ra A <3/4 (dpcm)
nho k cho minh voi nha
A=1/(1+3)+1(1+3+5)+1/(1+3+5+7)+....+1/(1+3+5+7+...+2017)
A=1/4+1/9+1/16+....+1/1018081
A=1/2^2+1/3^2+1/4^2+...+1/1009^2
Ta có : 1/3^2=1/3x3<1/2x3
1/4^2=1/4x4<1/3x4
......
1/1009^2<1/1008x1009
Suy ra 1/2^2+1/3^2+1/4^2+.....+1/1009^2<1/2^2+1/2x3+1/3x4+.....+1/1008x1009
Suy ra A< 1/2^2+1/2-1/3+1/3-1/4+.....+1/1008-1/1009
=> A<1/2^2+1/2+1/3-1/3+......+1/1008-1/1008-1/1009
=> A<1/2^2+( 1/2-1/1009)
=> A< 3023/4036
Mà +) 3023<3/4
+) A<3023/4026
Suy ra A<3/4
=> A<1008/1009
Ta có 1008/1009+
cho \(A=\frac{1}{1+3}+\frac{1}{1+3+5}+.....+\frac{1}{1+3+5+.....+2017}\)
chứng minh rằng: \(A< \frac{3}{4}\)
\(1+3+5+7+....+\left(2n+1\right)=\left\{\left[\left(2n+1\right)-1\right]:2+1\right\}.\frac{2n+2}{2}=\left(n+1\right)^2\)
Áp dụng ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}\)
Ta có :\(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1009^2}< \frac{1}{1008.1009}\)
\(\Rightarrow A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)
\(\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{1008}-\frac{1}{1009}=\frac{1}{4}+\frac{1}{2}-\frac{1}{1009}=\frac{3}{4}-\frac{1}{1009}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)
Cho \(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}.\)
Chứng minh rằng: \(A< \frac{3}{4}\)
Bài mình làm đơn giản thôi bạn nhé!
\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+7}+...+\frac{1}{1+3+5+..2017}\)
Ta có: \(\frac{1}{1+3}< \frac{3}{4}\)
\(\frac{1}{1+3+5}< \frac{3}{4}\)
\(\frac{1}{1+3+5+7}< \frac{3}{4}\)
. . . . . . . .
\(\frac{1}{1+3+5+...+2017}< \frac{3}{4}\)
____________________________________________________
\(A< \frac{3}{4}-\frac{1}{1+3+5+...+2017}\)
\(\Rightarrow A< \frac{3}{4}^{\left(đpcm\right)}\)
thằng tth quá ngu. làm vậy là sai bét.
hình như CTV mày câu và spam câu trả lời à
Nguyễn Phạm Nguyễn nói đúng oy. tth làm sai bét
Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(B=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{2017!}\)
Chứng minh \(A+B< 2\)
A=1/2^2 + 1/3^2 + 1/4^2 + ... + 1/2017^2
A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2016.2017
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2016 - 1/2017
A < 1 - 1/2017 < 1 (1)
B = 2!/3! + 2!/4! + 2!/5! + ... + 2!/2017!
B = 2!.(1/3! + 1/4! + 1/5! + ... + 1/2017!)
B < 2.(1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2016.2017)
B < 2.(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2016 - 1/2017)
B < 2.(1/2 - 1/2017) < 2.1/2 = 1 (2)
Từ (1) và (2) => A + B < 2 (đpcm)
chứng minh rằng:\(\frac{1}{5^3}+\frac{1}{6^3}+....+\frac{1}{2016^3}+\frac{1}{2017^3}< \frac{1}{40}\)
tuổi con HN là :
50 : ( 1 + 4 ) = 10 ( tuổi )
tuổi bố HN là :
50 - 10 = 40 ( tuổi )
hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi
ta có sơ đồ : bố : |----|----|----|
con : |----| hiệu 30 tuổi
tuổi con khi đó là :
30 : ( 3 - 1 ) = 15 ( tuổi )
số năm mà bố gấp 3 tuổi con là :
15 - 10 = 5 ( năm )
ĐS : 5 năm
mình nha
1.chứng minh rằng A<\(\frac{1}{16}\) biết A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+.....+\frac{99}{5^{100}}\)
2.tính (M-N)\(^3\) biết:
M=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
N=\(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2019}\)
\(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)
\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)
\(\Rightarrow4A=5A-A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)
Đặt \(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
Khi đó \(4A=B-\frac{99}{5^{100}}< B\)
\(5B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}\)
\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}+\frac{1}{5^{99}}\)
\(\Rightarrow4B=5B-B=1-\frac{1}{5^{99}}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{4\cdot5^{99}}< \frac{1}{4}\)
\(\Rightarrow4A < B\Rightarrow4A< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{16}\) ( đpcm )
2. \(M=\left(1+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(M=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
\(\Rightarrow\left(M-N\right)^3=0\)
Chứng minh A là một số nguyên dương :
A = \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+\)\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)^2+...+\left(\frac{1}{2017}\right)^2\)
cho A=\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2017}\)
\(A=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2017}\)
\(\Rightarrow A=\frac{1}{\frac{\left(3+1\right).\left[\left(3-1\right):2+1\right]}{2}}+\frac{1}{\frac{\left(5+1\right).\left[\left(5-1\right):2+1\right]}{2}}+...+\frac{1}{\frac{\left(2017+1\right).\left[\left(2017-1\right):2+1\right]}{2}}\)
\(\Rightarrow A=\frac{1}{\frac{4.2}{2}}+\frac{1}{\frac{6.3}{2}}+...+\frac{1}{\frac{2018.1009}{2}}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1009^2}\)
cho:
\(A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot.........\cdot\frac{2017}{2018}\)
chứng minh A<\(\frac{1}{2018}\)