CMR
\(\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+....\dfrac{2n-1}{4+\left(2n-1\right)^4}=\dfrac{n^2}{4n^2+1}\)
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22 tháng 11 2018
a) Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
Ta có:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n}+1\)
\(\Rightarrow1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 2-\dfrac{1}{n}\)
\(\Rightarrow\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{2}\right)\)
\(\Rightarrow A< \dfrac{1}{2^2}.2-\dfrac{1}{2^2}.\dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}-\dfrac{1}{2^3}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\left(Đpcm\right)\)
b) Đặt \(B=\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{\left(2n+1\right)^2}\)
Ta có:
\(B< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(B< \dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}\left(\dfrac{2n+1}{2n+1}-\dfrac{1}{2n+1}\right)\)
\(B< \dfrac{1}{2}.\dfrac{2n}{2n+1}\)
\(B< \dfrac{2n}{4n+2}\)
\(B< \dfrac{2n}{2\left(2n+1\right)}\)
\(B< \dfrac{n}{2n+1}\)
2 tháng 4 2022
2.
\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)
\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)
*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)
*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)
\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)
\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)
\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)
\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)
-Vậy \(n=1\)
2 tháng 4 2022
1. \(x^2+y^2=z^2\)
\(\Rightarrow x^2+y^2-z^2=0\)
\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)
-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.
\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.
-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.
*Xét \(\left(x-z\right)⋮2\):
\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.
*Xét \(\left(x+z\right)⋮2\):
\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
AH
Akai Haruma
Giáo viên
27 tháng 11 2023
1.
\(\lim (n^3+4n^2-1)=\infty\) khi $n\to \infty$
2.
\(\lim\limits_{n\to -\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to -\infty}\frac{-\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to -\infty}\frac{-(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{-1}{3}\)
\(\lim\limits_{n\to +\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to +\infty}\frac{\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to +\infty}\frac{(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{1}{3}\)
AH
Akai Haruma
Giáo viên
27 tháng 11 2023
3.
\(\lim \frac{1+2+...+n}{2n^2}=\lim \frac{n(n+1)}{4n^2}=\lim \frac{n^2+n}{4n^2}\\ =\lim (\frac{1}{4}+\frac{1}{4n})=\frac{1}{4}\)
4.
\(\lim \frac{3^n-4.2^{n-1}-10}{7.2^n+4^n}=\lim \frac{(\frac{3}{4})^n-(\frac{2}{4})^{n-1}-\frac{10}{4^n}}{7(\frac{2}{4})^n+1}\\ =\lim \frac{(\frac{3}{4})^n-(\frac{1}{2})^{n-1}-\frac{10}{4^n}}{7(\frac{1}{2})^n+1}\\ =\frac{0-0-0}{7.0+1}=0\)
HT
6 tháng 2 2021
\(a=\lim\limits\dfrac{3n^3-2n+1}{4n^4+2n+1}=\lim\limits\dfrac{\dfrac{3n^3}{n^4}-\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\dfrac{4n^4}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}=0\)
\(\Rightarrow\lim\limits\dfrac{-2n^2+1}{-n^2+3n+3}=\lim\limits\dfrac{-\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}+\dfrac{3}{n^2}}=-\dfrac{2}{-1}=2\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2021
\(a=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1+\dfrac{2}{n}}=\dfrac{0}{1}=0\)
\(b=\lim n^3\left(-2+\dfrac{1}{n}+\dfrac{2}{n^3}\right)=+\infty.\left(-2\right)=-\infty\)
\(c=\lim\dfrac{\sqrt{9-\dfrac{1}{n}-\dfrac{1}{n^2}}}{4-\dfrac{2}{n}}=\dfrac{\sqrt{9}}{4}=\dfrac{3}{4}\)
\(d=\lim\dfrac{\left(\dfrac{3}{4}\right)^n+5}{1+\left(\dfrac{2}{4}\right)^n}=\dfrac{5}{1}=5\)
Lời giải:
Ta có: \(4+(2n-1)^4=[(2n-1)^2+2]^2-[2(2n-1)]^2\)
\(=[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]\)
\(\Rightarrow \frac{2n-1}{4+(2n-1)^4}=\frac{2n-1}{[(2n-1)^2+2-2(2n-1)][(2n-1)^2+2+2(2n-1)]}\)
\(=\frac{1}{4}\left(\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)}\right)\)
Do đó:
\(\frac{1}{4+1^4}=\frac{1}{4}(1-\frac{1}{5})\)
\(\frac{3}{4+3^4}=\frac{1}{4}(\frac{1}{5}-\frac{1}{17})\)
\(\frac{5}{4+5^4}=\frac{1}{4}(\frac{1}{17}-\frac{1}{37})\)
......
Do đó:
\(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2n-1}{4+(2n-1)^4}=\frac{1}{4}(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{17}+...+\frac{1}{(2n-1)^2+2-2(2n-1)}-\frac{1}{(2n-1)^2+2+2(2n-1)})\)
\(=\frac{1}{4}(1-\frac{1}{(2n-1)^2+2+2(2n-1)})=\frac{1}{4}(1-\frac{1}{(2n-1+1)^2+1})\)
\(=\frac{1}{4}(1-\frac{1}{4n^2+1})=\frac{n^2}{4n^2+1}\)
Ta có đpcm.
n=1 ; \(\dfrac{1}{4+1^4}=\dfrac{1}{5}=\dfrac{1^2}{4.^2+1}=\dfrac{1}{5};dung\)
giả sử n =k đúng \(\Leftrightarrow S=\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}=\dfrac{k^2}{4k^2+1}\) (*)
cần c/m đúng n =k+1 ;
c/m
với n=k+1
\(S=\left(\dfrac{1}{4+1^4}+...+\dfrac{2k-1}{4+\left(2k-1\right)^4}\right)+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)
từ (*) =>\(S=\dfrac{k^2}{4k^2+1}+\dfrac{2\left(k+1\right)-1}{4+\left(2\left(k+1\right)-1\right)^4}\)
\(k+1=t\Leftrightarrow k=t-1\)
\(S=\dfrac{t^2-2t+1}{4\left(t^2-2t+1\right)+1}+\dfrac{2t-1}{4+\left(2t-1\right)^4}\)
\(S=\dfrac{t^2-2t+2}{4t^2-8t+5}+\dfrac{2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{\left(t^2-2t+1\right)\left(4t^2+1\right)+2t-1}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}\)\(S=\dfrac{t^2\left(4t^2-8t+5\right)}{\left(4t^2+1\right)\left(4t^2-8t+5\right)}=\dfrac{t^2}{\left(4t^2+1\right)}=\dfrac{\left(k+1\right)^2}{4\left(k+1\right)^2+1}\)
Vậy tổng trên đúng với k +1
theo Quy nạp ta có dpcm