b)

Để \(2n⋮\left(n-1\right)\)

\(\Rightarrow2.\left(n-1\right)+2⋮\left(n-1\right)\)

\(\Rightarrow2⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)\inƯ\left(2\right)=\left\{1;2\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=2\Rightarrow n=3\end{matrix}\right.\)

Vậy n=2;n=3 thì \(2n⋮\left(n-1\right)\)

c)

Để \(\left(3n-8\right)⋮\left(n-4\right)\)

\(\Rightarrow3.\left(n-4\right)+4⋮\left(n-4\right)\)

\(\Rightarrow4⋮\left(n-4\right)\)

\(\Rightarrow\left(n-4\right)\inƯ\left(4\right)=\left\{1;2;4\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}n-4=1\Rightarrow n=5\\n-4=2\Rightarrow n=6\\n-4=4\Rightarrow n=8\end{matrix}\right.\)

Vậy với .....................

@Luân Đào

Chứng minh đề bài sai

Ta có 

\(2^8+2=2\left(2^7+1\right)\)

=>\(A⋮2\)

A không chia hết cho 2 vì toàn bộ thừa số của A đều lẻ.

 t nghĩ đề là \(2^8+1\)

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n⁴ + 4 = (n²+2)² - 4n² = (n²-2n+2)(n²+2n+2) = [(n-1)²+1][(x+1)²+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

\(L=\lim\frac{1-2+4-\cdots+(-2)^{n-1}}{1-3+9-\ldots+(-3)^{n-1}}\)

Đặt \(A=1-2+4-\cdots+\left(-2\right)^{n-1}\)

\(=\left(-2\right)^0+\left(-2\right)^1+\cdots+\left(-2\right)^{n-1}\)

Số hạng đầu là \(u_1=1\)

Công bội là q=-2

Tổng của cấp số nhân A là:

\(A=1\cdot\frac{1 - (-2)^n}{1 - (-2)}=\frac{1 - (-2)^n}{3}\)

Đặt \(B=1-3+9-\cdots+\left(-3\right)^{n-1}\)

\(=\left(-3\right)^0+\left(-3\right)^1+\cdots+\left(-3\right)^{n-1}\)

Số hạng đầu là \(u_1=1\)

Công bội là q=-3

Tổng của cấp số nhân B là:

\(B=1\cdot\frac{1 - (-3)^n}{1 - (-3)}=\frac{1 - (-3)^n}{4}\)

\(\frac{A}{B}=\frac{\frac{1 - (-2)^n}{3}}{\frac{1 - (-3)^n}{4}}=\frac{4\left[1 - (-2)^n\right]}{3\left[1 - (-3)^n\right]}\)

\(L=\lim\frac{1-2+4-\cdots+(-2)^{n-1}}{1-3+9-\ldots+(-3)^{n-1}}\)

\(=\lim\frac{4}{3}\cdot\frac{1 - (-2)^n}{1 - (-3)^n}\)

\(=\lim\frac{4}{3}\cdot\frac{\frac{1}{(-3)^n} - \left(\frac{-2}{-3}\right)^n}{\frac{1}{(-3)^n} - 1}\)

\(=\lim\frac{4}{3}\cdot\frac{\frac{1}{(-3)^n} - \left(\frac{2}{3}\right)^n}{\frac{1}{(-3)^n} - 1}\)

\(=\frac{4}{3}\cdot\frac{0 - 0}{0 - 1}=\frac{4}{3}\cdot0=0\)

a) \(\dfrac{32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=8=\left(-2\right)^3\)

=> n = 3

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4=2^2\)

=> n = 2

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)

=> 2n - 1 = 3

=> 2n = 4

=> n = 2

Giải:

a) \(\dfrac{32}{\left(-2\right)^n}=4\) 

\(\Rightarrow\left(-2\right)^n=32:4=8\) 

\(\Rightarrow\left(-2\right)^n=8\) 

Vì \(\left(-2\right)^n=2^3\) là ko thể nên n ∈ ∅

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=8:2=4\) 

\(\Rightarrow2^n=4\) 

\(\Rightarrow2^n=2^2\) 

\(\Rightarrow n=2\) 

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\) 

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\) 

\(\Rightarrow2n-1=3\rightarrow n=2\)