1/4^2 + 1/5^2 + 1/6^2 + ... + 1/64^2 < 5/16
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11 tháng 6 2019
Đặt \(A=\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{64^2}\)
Đặt \(B=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}\)
Ta có: \(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
....................
\(\frac{1}{64^2}< \frac{1}{63.64}\)
\(\Rightarrow B< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)
\(\Rightarrow B< \frac{1}{4}-\frac{1}{64}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
\(\Rightarrow A< \frac{1}{4^2}+\frac{1}{4}\)
\(\Rightarrow A< \frac{5}{16}\)
XO
11 tháng 6 2019
Ta có S =\(\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{64^2}\)
= \(\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{64.64}\)
< \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)
= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{63}-\frac{1}{64}\)
= \(\frac{1}{3}-\frac{1}{64}\)
= \(\frac{61}{192}\)> \(\frac{60}{192}=\frac{5}{16}\)
S < \(\frac{61}{192}>\frac{5}{16}\)
=> sai đề
26 tháng 6 2023
15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9
Sửa đề:
A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007
=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007
=-1+1+1/2007=1/2007
16:
\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)
=1/64
17:
=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7
=-6/7
8 tháng 5 2019
A=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
=>A<\(\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\)
=>A<\(\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)\(:2\)=\(\left(\frac{1}{2}-\frac{1}{14}\right):2\)<\(\frac{1}{2}\)
=>A<\(\frac{1}{2}\)
Ta có: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}< \frac{1}{4^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)
\(\frac{1}{4^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}=\frac{1}{4^2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{63}-\frac{1}{64}\)
\(=\frac{1}{4^2}+\frac{1}{4}-\frac{1}{64}\)
VÌ: \(\frac{1}{4^2}+\frac{1}{4}-\frac{1}{64}< \frac{1}{4^2}+\frac{1}{4}=\frac{5}{16}\)
Nên: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}< \frac{5}{16}\left(dpcm\right)\)