chung minh rang:1/12<1/2^3+1/3^3+...+1/n^3+...+1/2017^3<505/2018(voi moi n>1)
Những câu hỏi liên quan
Chung minh rang : 1/11+1/12+1/13+...+1/50 < 8/5
chung minh rang 11^n+2+12^2n+1 chia het cho 133
chung minh rang A=(17^n+1)(17^n+2)chia het cho 3 voi moi n thuoc N
cho (2a+7b) chia het cho 3 ( a b thuoc N). chung to (4a+2b) chia het cho 3
chung minh rang A=1/11×1/12×1/13+...+1/50>1/3
chung minh rang
a)1/101+1/102+1/103+...+1/200<7/12
Cho S= 3/1+3/11+3/12+3/13+3/14
Chung minh rang 1<S<2
* S = 3/10+3/11+3/12+3/13+3/14 < 3/10 + 3/10+3/10+3/10+3/10
< 3/10 x 5
< 3/2 < 2sư
* S = 3/10+3/11+3/12+3/13+3/14 > 3/15+3/15+3/15+3/15+3/15
> 3/15 x 5
> 1
CHỨNG TỎ ........
> 1
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cho s=1 phan 11+1 phan 12+1phan13+.....+1phan19+1phan20
chung minh rang 1phan2 lon hon S lon hon 1
cho A=1/101+1/102+...+1/200
chung minh rang
a) A>7/12
b)A>5/8
câu b
C= 1/181+1/182+...1/200< 20/200=1/10
A=B+C<4/9+1/10=40/90+9/90=49/90 mà 49/90<3/4 ( quy đồng)
Vậy A<3/4
** D= 1/101+1/101+...1/150>50.(1/101)=50/101>1...
E= 1/151+1/152+...+1/200> 50.(1/151)=50/151>1/3
D+E>1/3+1/3=2/3 mà 2/3>5/8
Vậy A>5/8
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a)Ta CM: S(n)>7/12 (*) bằng qui nạp
+S(3)=1/4+1/5+1/6>7/12
+giã sử S(k)>7/12 (k>=3, k nguyên)
tức là:S(k)=1/(k+1)+1/(k+2)+...+1/2k>7/12
+Ta có: S(k+1)=1/(k+2)+1/(k+3)+...+1/(2k+2)
=1/(k+1)+1/(k+2)+...
..+1/2k+1/(2k+1)+1/(2k+2)-1/(k+1)
=S(k)+1/(2k+1)+1/(2k+2)-1/(k+1)
=S(k)+1/[(2k+1)(2k+2)]>7/2
theo nguyên lí qui nạp=>(*) đúng với mọi n>3, n nguyên
câu b tương tự
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Chung minh rang
12 8 . 912 = 1816
1 . chung minh rang mot so tu nhien n khong the dong thoi (7n-1)chia het cho 4 va (5n+3) chia het cho 12