Tìm n biết \(2^2\cdot3^{2\left(n-1\right)}=\left(2n\right)^2\)
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\(a=\lim\dfrac{\left(n-2\right)!\left(n-1+\left(n-1\right)n\right)}{\left(n-2\right)!\left(\left(n+2\right)\left(n+1\right)n\left(n-1\right)-1\right)}+\lim\dfrac{3}{\left(n+2\right)!-\left(n-2\right)!}\)
\(=\lim\dfrac{n^2-1}{\left(n+2\right)\left(n+1\right)n\left(n-1\right)-1}+\lim\dfrac{3}{\left(n+2\right)!-\left(n-2\right)!}\)
\(=0+0=0\)
\(b=\lim\dfrac{2+\dfrac{1}{n}}{3^n}=\dfrac{2}{\infty}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
$\frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}(*)$
Với $n=1$ thì $(*)\Leftrightarrow \frac{1}{2}=\frac{1}{2}$
Vậy $(*)$ đúng với $n=1$
Giả sử với $n=k$,$ k\in \mathbb{N^*}$ thì $(*)$ đúng, tức là:
$\frac{1.3.5...(2k-1)}{(k+1)(k+2)...(k+k)}=\frac{1}{2^k}$
Ta cần chứng minh với $n=k+1$ thì $(*)$ đúng, tức là:
$\frac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}=\frac{1}{2^{k+1}}=\frac{1}{2^k}.\frac{1}{2}$
$\Leftrightarrow \frac{1.3.5...(2k+1)}{(k+2)(k+3)...(2k+2)}=\frac{1.3.5...(2k-1)}{2(k+1)(k+2)...(k+k)}$
$\Leftrightarrow \frac{1.3.5...(2k-1)2k(2k+1)}{(k+2)(k+3)...2k(2k+1)(2k+2)}=\frac{1.3.5...(2k-1)}{2(k+1)(k+2)...2k}$
$\Leftrightarrow \frac{2k(2k+1)}{2k(2k+1)(2k+2)}=\frac{1}{2(k+1)}$
$\Leftrightarrow \frac{1}{(2k+2)}=\frac{1}{2(k+1)}$
Do đó với $n=k+1$ thì $(*)$ đúng
$\Rightarrow \frac{1.3.5...(2n-1)}{(n+1)(n+2)...(n+n)}=\frac{1}{2^n}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)
b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)
c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)
d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)
e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)
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A = 1.2.3 + 2.3.4 + 3.4.5 ... + n(n + 1)(n + 2)
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ... + n(n + 1)(n + 2).4
4A = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2)+ ... + n(n + 1)(n + 2)[(n + 3) - (n - 1)]
4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + n(n + 1)(n + 2)(n + 3) - (n-1)n(n+1)(n+2)
4A = n(n+1)(n+2)(n+3)
A = n(n + 1)(n+2)(n + 3) : 4