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\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow S_n=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow S_n=1-\frac{1}{\sqrt{n+1}}\)
\(lim\left(S_n\right)=lim\left(1-\frac{1}{\sqrt{n+1}}\right)=1-0=1\)
S là tổng cấp số nhân vô hạn với \(\left\{{}\begin{matrix}u_1=1\\q=-\frac{1}{3}\end{matrix}\right.\)
Theo công thức ta có: \(S=\frac{u_1}{1-q}=\frac{1}{1-\left(-\frac{1}{3}\right)}=\frac{3}{4}\)
a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
Câu 1:
$S=1+\cos ^2x+\cos ^4x+...+\cos ^{2n}x=1+\cos ^2x+(\cos ^2x)^2+...+(\cos ^2x)^n=\frac{(\cos ^2x-1)(1+\cos ^2x+(\cos ^2x)^2+...+(\cos ^2x)^n}{\cos ^2x-1}$
$=\frac{(\cos ^2x)^{n+1}-1}{\cos ^2x-1}=\frac{\cos ^{2n+2}x-1}{\sin ^2x}$
Đặt \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}=A\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)
\(\Leftrightarrow A=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
\(A=lim\frac{\sqrt{n+2}+\sqrt{n+1}}{1}=lim\left[n\left(\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}\right)\right]=+\infty.2=+\infty\)
\(B=lim\frac{8^3.64^n-9.27^n}{4^4.64^n+5^3.25^n}=\frac{8^3-9.\left(\frac{27}{64}\right)^n}{4^4+5^3\left(\frac{25}{64}\right)^n}=\frac{8^3}{4^4}=2\)
\(1;-\frac{1}{2};\frac{1}{4}...\) là dãy cấp số nhân lùi vô hạn có \(u_1=1\) và \(q=-\frac{1}{2}\)
Do \(\left|q\right|< 1\) nên theo công thức tổng cấp số nhân:
\(S_n=\frac{u_1}{1-q}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\)
Câu 4.
\(\lim \left( {{n^2}\sin \dfrac{{n\pi }}{5} - 2{n^3}} \right) = \lim {n^3}\left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - \infty \)
Vì \(\lim {n^3} = + \infty ;\lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2 \)
\(\left| {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n}} \right| \le \dfrac{1}{n};\lim \dfrac{1}{n} = 0 \Rightarrow \lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2\)
Câu 5.
Ta có: \(\left\{ \begin{array}{l} 0 \le \left| {{u_n}} \right| \le \dfrac{1}{{{n^2} + 1}} \le \dfrac{1}{n} \to 0\\ 0 \le \left| {{v_n}} \right| \le \dfrac{1}{{{n^2} + 2}} \le \dfrac{1}{n} \to 0 \end{array} \right. \to \lim {u_n} = \lim {v_n} = 0 \to \lim \left( {{u_n} + {v_n}} \right) = 0\)
Lời giải:
\(S_{n}=\frac{1}{3^1}-\frac{1}{3^2}+....+\frac{(-1)^{n+1}}{3^n}\)
\(3S_n=1-\frac{1}{3}+....+\frac{(-1)^{n+1}}{3^{n-1}}\)
Cộng theo vế:
\(4S_n=1+\frac{(-1)^{n+1}}{3^n}=1-\left(\frac{-1}{3}\right)^n\)
\(\lim(S_n)=\frac{\lim(4S_n)}{4}=\frac{1}{4}\lim [1-\left(\frac{-1}{3}\right)^n]=\frac{1}{4}\) (nhớ rằng \(\lim\limits q^n=0\) với $|q|< 1$)
Đáp án A.