A=1/101+1/102+1/103+...1/200 chung minh rang 5/8<A<3/4
Những câu hỏi liên quan
chung minh rang
a)1/101+1/102+1/103+...+1/200<7/12
cho A =1/101+1/102+1/103+...+1/200
CHUNG TO RANG
A)A>7/12
B) A>5/8
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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cho A=1/101+1/102+1/103+...+1/200 Chứng minh rằng A > 5/8
chung minh rang :1/101+1/102+1/103+...+1/400<11/6
A=1/100+1/101+1/102+1/103+...+1/200 chưng minh a>5/8
chung minh rang
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
giup minh minh like cho nho giai chi tiet mot chut nhe
1-1/2+1/3-1/4+...+1/199-1/200=(1+1/2+1/3+1/4+...+199+1/200)-(1+1/2+1/3+...+1/100)=1+1/2+1/3+1/4+...+1/199+1/200-1-1/2-1/3-1/4-...-1/99-1/100=(1+1/2+1/3+...+1/100)-(1+1/2+1/3+...+1/100)+(1/101+1/102+...+1/200)=0+(1/101+1/102+...+1/200)=(1/101+1/102+...+1/200)(đpcm)
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Chung minh rang:
a,1-1/2+1/3-1/4+...+1/199-1/200=1/101+1/102+...+1/200
Chứng minh: A > 7/12 và A > 5/8 với A = 1/101 + 1/102 + 1/103 + ... + 1/200