Cho A =3/5^3+4/5^4+5/5^5+......+102/5^102+103/5^103. Chứng minh rằng : A<13/100
Cho A= 6/101+6/102+6/103+...+6/199+6/200. Chứng minh rằng 3<A<5
cho A=1/101+1/102+1/103+...+1/200 Chứng minh rằng A > 5/8
Giúp mình 5 câu này nhé . Ai làm đc cả 5 câu cho 10 điểm luôn ( Nếu đúng )
1/Cho A= 1/101+1/102+1/103+...+1/150
a) So sánh 1/150 với 1/101;...; 1/150 với 1/149 <----------------KO PHẢI LÀM
b) Chứng minh : A > 1/3
2/ Cho A= 1/101+1/102+1/103+...+1/200
a) So sánh: 1/101+1/102+...+1/150với 1/3 và 1/151+1/152+...+1/200 với 1/4
b) Chứng minh: A > 7/12
3/Cho A= 1/101+1/102+...+1/200
Chứng minh: 1/2 < A < 1
4/ Cho A = 1/101+1/102+1/103+...+1/150. Chứng minh: 1/3 < A < 1/2
5/ Chứng minh: 1/5+1/14+1/28 < 1/3
CHÚC CÁC BẠN THÀNH CÔNG
CÁC BẠN CHỈ CẦN GIÚP MÌNH ÍT NHẤT 2 CÂU THÔI
1 Chứng minh
a) A= (-1/5)^0 + (-1/5)^1 + (-1/5)^2 + ..........+(-1/5)^2014 < 5/6
b) B= 1/101 + 1/102 + 1/103 + ....+ 1/200 < 3/4
Giải hộ mình với mai phải nộp euif
cho A= 1+2-3+4-5+6+...+100+101-102+103
B= 1+(-3)+5+(-7)+...+101+(-103)+105
so sánh A và B
A= [(1+101)x101:2]-(102-103)
A= 5151+1
A=5152
B= [1+(-3)]+[4+(-5)]+.......[101+(-103)]+105
B= (-2)+(-2)...........+(-2)+105
=> A>B
B=(-2)x26+105
B=(-56)+105
B= 49
B=5/3(1−1/4+1/4−1/7+...+1/100−1/103)=5/3(102/103)=170103 ai giả thích hộ mình lấy 5/3 ở đâu không
THAM KHẢO:
Ta có: 5/1x4 + 5/4x7 + ... + 5/100x103
= 5/3 x (1/1 - 1/4 + 1/4 - 1/7 +...+1/100 - 1/103)
= 5/3 x (1 - 1/103)
= 5/3 x 102/103
= 170/103
chứng minh rằng:
a) A= 1/5 + 1/45 + 1/117 + 1/221 + 1/357 + 1/525 < 1/4
b) B= 1/101 + 1/102 + 1/103 +...+ 1/200 < 5/8
a) \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+\frac{1}{221}+\frac{1}{357}+\frac{1}{525}\)
\(\Rightarrow A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\)
\(\Rightarrow4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{21.25}\)
\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{25}\)
\(4A=\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow A=\frac{24}{25}\div4=\frac{6}{25}
A=1/101+1/102+1/103+...1/200 chung minh rang 5/8<A<3/4
cho A=1/101+1/102+1/103+...+1/199+1/200 cmr 5/8<a<3/4
\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)
\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)
\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)
\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)
\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)
\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)
So sánh A và B biết:
A-1/2*3/4*5/6*...*101/102
B=2/3*3/4*...*102/103