A=1/3^2+1/5^2+1/7^2+...+1/2019^2 chung minh A<1/4
a)chung minh A= 2^1+2^2+2^3+2^4+...2^2010chia het cho 3
b)chung minh B= 3^1+3^2+3^3+3^4+...3^2010chia het cho 4
c)chung minh C= 5^1+5^2+5^3+5^4+...5^2010chia het cho 6
d)chung minh D= 7^1+7^2+7^3+7^4+...7^2010chia het cho 8
a) A=21+22+23+...+22010
A=(21+22)+(23+24)+.....+(22009+22010)
A=(21x3)+(23x3)+.....+(22009x3)
A=3x(21+23+.......+22009)
Vậy A chia hết cho 3.
NHỮNG CÂU CÒN LẠI BẠN LÀM TƯƠNG TỰ !
Tính giá trị của biểu thức A=(3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+...+(2019^2+1)/(2019^2-1)
Cho A=\(\frac{1}{3^2}+\frac{1}{5^5}+\frac{1}{7^2}+...+\frac{1}{2019^2}\)
Chứng minh A<\(\frac{1}{2}\)
Ta có :
\(A=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2019^2}< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
Cho \(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(\Rightarrow S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(\Leftrightarrow S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(\Leftrightarrow S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1009}{4040}< \frac{1}{2}\)
Mà A < S ⇒ đpcm
Cho A = \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+.....+\dfrac{1}{2019^2}\)
Chứng minh rằng \(\dfrac{20}{101}< A< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)
A=1+3+3^2+3^3+3^4+3^5+3^6+3^7 chung minh A=(3^8-1):2
\(A=1+3+...+3^7\\ \Rightarrow3A=3+3^2+...+3^8\\ \Rightarrow3A-A=\left(3+3^2+...+3^8\right)-\left(1+3+...+3^7\right)\\ \Rightarrow2A=3^8-1\\ \Rightarrow A=\dfrac{3^8-1}{2}\)
\(A=1+3+3^2+...+3^7\)
\(3A=3+3^2+3^3+...+3^8\)
\(3A-A=\left(3+3^2+3^3+...+3^8\right)-\left(1+3+3^2+...+3^7\right)\)
\(2A=3^8-1\)
\(\Rightarrow A=\dfrac{3^8-1}{2}\) (đpcm)
cho A =1/2*3/4*5/6*...*99/100
B=2/3*4/5*6/7*...*100/101
C=1/2*2/3*4/5*...*98/99
a) so sanh A, B, C
b) Chung minh: A*C< A^2< 1/10
c) Chung minh: 1/15< A< 1/10
Lam giup minh di ai lam duoc minh tich dung cho
chung minh A= 2 + 2^2 +2^3 +2^4 +.........+2^60 chia het cho 7
tim so tu nhien n de : n+4 chia het cho n+1
chung minh ( 1+2 +2^2 +2^3+2^4+2^5+2^6+2^7) chia het cho 3
1. A = 2 + 22 + 23 + 24 + ... + 260
A = ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + ... + ( 258 + 259 + 260 )
A = 2 ( 1 + 2 + 22 ) + 24 ( 1 + 2 + 22 ) + ... + 258 ( 1 + 2 + 22 )
A = 2 . 7 + 24 . 7 + ... + 258 . 7
A = ( 2 + 24 + ... + 258 ) . 7 => A \(⋮\)7
Vậy ...
2.Ta có : \(n+4⋮n+1\)
Mà : \(n+1⋮n+1\)
\(\Rightarrow\left(n+4\right)-\left(n+1\right)⋮n+1\Rightarrow n+4-n-1⋮n+1\)
\(\Rightarrow3⋮n+1\Rightarrow n+1\in\left\{1;3\right\}\)
\(\Rightarrow n\in\left\{0;2\right\}\)
3. Đặt B = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
B = ( 1 + 2 ) + ( 22 + 23 ) + ( 24 + 25 ) + ( 26 + 27 )
B = ( 1 + 2 ) + 22 ( 1 + 2 ) + 24 ( 1 + 2 ) + 26 ( 1 + 2 )
B = 1 . 3 + 22 . 3 + 24 . 3 + 26 . 3
B = ( 1 + 22 + 24 + 26 ) . 3 \(\Rightarrow\) B \(⋮\)3
Vậy ...
cho a1/a2=a2/a3=a3/a4=...=a2018/a2019.chung minh:
a1/a2=(a1+a2+a3+...+a2018/a2+a3+a4+...+a2019)2018
áp dụng t/c dãy tỉ số bằng nhau, ta có:
\(\frac{a1}{a2}=\frac{a2}{a3}=....=\frac{a2018}{a2019}\)
\(=\frac{a1+a2+..+a2018}{a2+a3+..+a2019}\)
\(=>\frac{a1}{a2}=\frac{a1+a2+..+a2018}{a2+a3+..+a2019}\left(dpcm\right)\)
1.Cho A=2020/20192+1 +2020/20192+2 +...+2020/20192+2019
Chứng minh A không thuộc N
2. Tìm a,b,c thuộc N sao
a) 1/a+1/b=7
b)1/a+1/b+1/c=2
c) 1/a + 1/a+b +1/a+b+c =1