tìm n biết :\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)=2010
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HT
0
NN
1
10 tháng 1 2017
Áp dụng \(1+2+...+k=\frac{k\left(k+1\right)}{2}\) thì ta được :
\(\sqrt{\left[1+2+3+...+\left(n-1\right)+n\right]+\left[n+\left(n-1\right)+...+3+2+1\right]-n}=2010\)
\(\Leftrightarrow\sqrt{2.\frac{n\left(n+1\right)}{2}-n}=2010\)
\(\Leftrightarrow\sqrt{n^2}=2010\Leftrightarrow n=2010\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)
\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)
\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)
\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
\(=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{2.1}{1.1}=2\)
DT
2
30 tháng 8 2016
Vì \(n\in Z^+\)nên\(n\left(n+1\right)\left(n+2\right)>n^3\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)}>n\)
\(\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}>n\)(1)
Lại có:\(n^2+2n+1>n^2+2n\Rightarrow\left(n+1\right)^2>n\left(n+2\right)\Rightarrow\left(n+1\right)^3>n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)}\\ \Rightarrow\sqrt[3]{n^3+3n^2+3n+1}>\sqrt[3]{n^3+3n^2+2n}\)
\(\Rightarrow\sqrt[3]{n^3+3n^2+2n+n+1}>\sqrt[3]{n^3+3n^2+2n+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)
\(\Rightarrow\sqrt[3]{\left(n+1\right)^3}>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)
Tương tự \(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)(2)
Từ (1) và (2) suy ra:
\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< n+1\)
30 tháng 8 2016
\(n\in Z^+\)nên n2 < n2 + 2n < n2 + 2n + 1 <=> n2 < n(n + 2) < (n + 1)2 => n3 < n(n + 1)(n + 2) < (n + 1)3
=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< n+1\)
=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n+1}\)\(=\sqrt[3]{\left(n+1\right)\left(n^2+2n+1\right)}=\sqrt[3]{\left(n+1\right)\left(n+1\right)^2}=n+1\)
=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)
\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}}< n+1\)
Tiếp tục như vậy,ta có đpcm.
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
25 tháng 11 2023
1: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+n^2+n+1-n^3}{\sqrt[3]{\left(n^3+n^2+n+1\right)^2}+n\cdot\sqrt[3]{n^3+n^2+n+1}+n^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+1}\)
\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)
2: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}\)
\(=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 2 2021
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)
Xét 1+2+3+...+(n-1)
Tổng trên có số số hạng là:
(n-1-1):1+1 = n-1 (số)
Tổng trên là:
\(\frac{\left(n+1-1\right)\left(n-1\right)}{2}=\frac{n\left(n-1\right)}{2}\)
=> Thay vào, ta có:
\(\sqrt{\frac{n\left(n-1\right)}{2}+n+\frac{n\left(n-1\right)}{2}}=2010\)
=> \(\sqrt{n\left(n-1\right)+n}=2010\)
=> \(\sqrt{n\left(n-1+1\right)}=2010\)
=> \(\sqrt{n.n}=2010\Rightarrow\sqrt{n^2}=2010\)
=> n = 2010
Bạn áp dụng đáp án phía dưới vào.
Có:
\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+....+3+2+1}=n\)(Tính ở câu dưới)
Mà \(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+....+3+2+1}=2010\)(Đề bài)
=> n = 2010