chung minh rang :1/101+1/102+1/103+...+1/400<11/6
Những câu hỏi liên quan
A=1/101+1/102+1/103+...1/200 chung minh rang 5/8<A<3/4
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
chung minh : A=1/101+1/102+1/103+.....+1/150>1/3
Ta thấy tổng trên có 50 số hạng .
Ta có:
1/101>1/150
1/102>1/150
...
1/149>1/150
1/150=1/150
=>1/101+1/102+...+1/149+1/150>1/150+1/150+...+1/150
---50 số hạng 1/150-------
=>1/101+1/102+...+1/149+1/150>1/150.50
=>1/101+1/102+...+1/149+1/150>50/150
=>1/101+1/102+...+1/149+1/150>1/3
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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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cho s biet:1/101+1/102+....+1/130. chung minh rang 1/4<s<91/330
S=(1/101+1/102+...+1/110)+(1+111+...+1/120)+(1/121+...+1/130)
=>1/110.10+1/120.10+1/130.10=1/11+1/12+1/13>1/12+2/12=1/4 (dễ có :
1/11+1/13>2/12
=>S>1/4(1)
+)S=1/101+1/130)+(1/102+1/129)+......+(1/115+1/116)(có 15 cặp
=231/101.130+231/102.129+...231/115.116=231
(1/101.130+1/102.129+...+1/115.116)
Ta có nhận xét tích 101 .130 có giá trị nhỏ nhất ,thật vậy :
xét 102.129=(101+1).(130-1)=101.130-101+130-1=101.130+28>101.130
Tương tự các cặp cong lại ,ta có : 1/101.130+1/129.102+....+1/115.116<1/101.130.15
=>S=231.1/101.130.15=693/2626<91/330
từ (1)(2)=>đpcm
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S=(1/101+1/102+...+1/110)+(1+111+...+1/120)+(1/121+...+1/130)
=>1/110.10+1/120.10+1/130.10=1/11+1/12+1/13>1/12+2/12=1/4 (dễ có :
1/11+1/13>2/12
=>S>1/4(1)
+)S=1/101+1/130)+(1/102+1/129)+......+(1/115+1/116)(có 15 cặp
=231/101.130+231/102.129+...231/115.116=231
(1/101.130+1/102.129+...+1/115.116)
Ta có nhận xét tích 101 .130 có giá trị nhỏ nhất ,thật vậy :
xét 102.129=(101+1).(130-1)=101.130-101+130-1=101.130+28>101.130
Tương tự các cặp cong lại ,ta có : 1/101.130+1/129.102+....+1/115.116<1/101.130.15
=>S=231.1/101.130.15=693/2626<91/330
từ (1)(2)=>đpcm
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chung minh rang 1-1/2+1/3-1/4+...+1/99-1/200=1/101+1/102+...+1/199+1/200
chung minh rang 1-1/2+1/3-1/4+.....+1/199-1/200=1/101+1/102+.....+1/200
bien doi so sdau tien
1-1/2+1/3-1/4+...+1/199-1/200
(1+1/3+1/5+...+1/199)-(1/2+1/4+1/6+...+1/200)
(1+1/2+1/3+1/4+1/5+1/6+...+1/199+1/200)-2*(1/2+1/4+1/6+,,,+1/200)
(1+1/2+1/3+1/4+...+1/200)-((1+1/2+1/3+...+1/100)
1/101+1/102+1/103+...+1/200=so bewn canh
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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