Với n là số tự nhiên. Hãy so sánh:
[n^2+n+2;3] và n^2+n+2
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YN
9 tháng 2 2023
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
ND
2
MT
16 tháng 7 2015
\(\text{a)}A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}
K
6 tháng 8 2018
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TM
11 tháng 5 2016
ta có: \(\frac{n}{n+3}=\frac{n\left(n+2\right)}{\left(n+3\right)\left(n+2\right)}=\frac{n^2+2n}{\left(n+3\right)\left(n+2\right)}\)
\(\frac{n+1}{n+2}=\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+3\right)}=\frac{n^2+3n+n+3}{\left(n+2\right)\left(n+3\right)}\)
thấy rõ \(\frac{n^2+2n}{\left(n+3\right)\left(n+2\right)}<\frac{n^2+3n+n+3}{\left(n+3\right)\left(n+2\right)}\Rightarrow\frac{n}{n+3}<\frac{n+1}{n+2}\)
Ngoài ra bạn có thể sử dụng phương pháp so sánh phần bù