Cho A = 1/32+1/42+1/52+.....+1/502 . Chứng minh rằng :
a) A > 1/4
b) A < 4/9
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`Answer:`
\(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)
a) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)
\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)
b) Ta thấy:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)
\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)
\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)
\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)
\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)
giả sử \(a_1\left(1-a_2\right);a_2\left(1-a_3\right);...;a_9\left(1-a_1\right)>\frac{1}{4}\)
\(\Rightarrow a_1\left(1-a_2\right).a_2\left(1-a_3\right)...a_9\left(1-a_1\right)>\left(\frac{1}{4}\right)^9\)
mà\(a_1\left(1-a_1\right)=a_1-a^2_1=\frac{1}{4}-\left(\frac{1}{2}-a_1\right)^2\le\frac{1}{4}\)
CMTT \(a_2\left(1-a_2\right);a_3\left(1-a_3\right);...;a_9\left(1-a_9\right)\le\frac{1}{4}\)
=> gt sai=>phải có 1hs bé hơn 1/4
ta có:\(a< b\Rightarrow4a< 4b\) và \(1< 3\)
\(\Rightarrow4a+1< 4b+3\)
Câu b tương tự nhưng nhớ đổi dấu khi nhân vs số âm
Ta có
\(A>\frac{1}{3^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}\)
\(\Rightarrow A>\frac{1}{9}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{50}-\frac{1}{51}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{9}-\frac{1}{51}\right)\)
\(\Rightarrow A>\frac{1}{4}+\frac{42}{9.51}>\frac{1}{4}\)
Vậy A>1/4
b)
Ta có
\(A< \frac{1}{3}^2+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{9}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{50}\)
\(\Rightarrow A< \frac{4}{9}-\frac{1}{50}< \frac{4}{9}\)
Vậy A<4/9
thank nha