3^4+8^9+15/16+...+1599/1600 lớn hơn 38
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Ta có : \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{1599}{1600}\)
\(=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{1600}\right)\)
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)
\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)...\left(1-\frac{1}{1601}\right)\)
Vì \(\frac{1}{2}>\frac{1}{3};\frac{1}{4}>\frac{1}{5};\frac{1}{6}>\frac{1}{7};...;\frac{1}{1600}>\frac{1}{1601}\)
\(\Rightarrow1-\frac{1}{2}< 1-\frac{1}{3};1-\frac{1}{4}< 1-\frac{1}{5};1-\frac{1}{6}< 1-\frac{1}{7};...;1-\frac{1}{1600}< 1-\frac{1}{1601}\)
\(\Rightarrow A< B\)
hay A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\)
Vậy A<\(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta luôn có:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{7}< \frac{6}{7}\)
\(........\)
\(\frac{1599}{1600}< \frac{1600}{1601}\)
Từ trên: \(\Rightarrow A=\frac{1}{2}.\frac{3}{4}....\frac{1599}{1600}\left(1\right)\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}...\frac{1599}{1600}< \frac{2}{3}.\frac{4}{5}....\frac{1600}{1601}\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{1600}{1601}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(\frac{1}{2}< \frac{2}{3},\frac{3}{4}< \frac{4}{5},...,\frac{1599}{1600}< \frac{1600}{1601}\)
Do đó ta có
A=\(\frac{1}{2}\times\frac{3}{4}\times...\times\frac{1599}{1600}< \frac{2}{3}\times\frac{4}{5}\times...\times\frac{1600}{1601}\)
#Châu's ngốc
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