\(p=\left(n-1\right)^2\left[\left(n-1\right)^2+1\right]+1\)

\(\left(n-1\right)^4+2.\left(n-1\right)^2+1-\left(n-1\right)^2\)

\(\left[\left(n-1\right)^2+1\right]^2-\left(n-1\right)^2\)

\(\left[\left(n-1\right)^2+1-\left(n-1\right)\right]\left[\left(n-1\right)^2+1+\left(n-1\right)\right]\)

\(\left[n^2-3n+3\right]\left[n^2-n+1\right]\)

can

\(\orbr{\begin{cases}n^2-3n+3=1\Rightarrow n=\orbr{\begin{cases}n=2\\n=1\end{cases}}\\n^2-n+1=1\Rightarrow n=\orbr{\begin{cases}n=0\\n=1\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}n^2-3n+3=1\\n^2-n+1=1\end{cases}}\)

n=(0,1,2)

du

n=2

ds: n=2

Lời giải:

$A=\frac{2}{3}+\frac{4}{3^2}+\frac{6}{3^3}+...+\frac{2n}{3^n}$

$3A=2+\frac{4}{3}+\frac{6}{3^2}+....+\frac{2n}{3^{n-1}}$

$3A-A=2+\frac{2}{3}+\frac{2}{3^2}+....+\frac{2}{3^{n-1}}-\frac{2n}{3^n}$

$2A=2+\frac{2}{3}+\frac{2}{3^2}+....+\frac{2}{3^{n-1}}-\frac{2n}{3^n}$

$A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}-\frac{n}{3^n}$

$3A=3+1+\frac{1}{3}+....+\frac{1}{3^{n-2}}-\frac{n}{3^{n-1}}$

$3A-A=3-\frac{1}{3^{n-1}}-\frac{n}{3^{n-1}}+\frac{n}{3^n}$

$2A=3-\frac{n+1}{3^{n-1}}+\frac{n}{3^n}$

$2A=\frac{3^{n+1}-2n-3}{3^n}$

$A=\frac{3.3^n-2n-3}{2.3^n}$

$\Rightarrow a=3; b=1; c=2\Rightarrow abc=6$

Ta có: \(B=\left(n^2-8\right)^2+36\)

\(=n^4-16n^2+64+36\)

\(=n^4+20n^2+100-36n^2\)

\(=\left(n^2+10\right)^2-\left(6n\right)^2=\left(n^2-6n+10\right)\left(n^2+6n+10\right)\)

Để B là số nguyên tố thì một trong hai thừa số \(n^2-6n+10;n^2+6n+10\) phải bằng 1 và thừa số còn lại là số nguyên tố

TH1: \(n^2-6n+10=1\)

=>\(n^2-6n+9=0\)

=>\(\left(n-3\right)^2=0\)

=>n-3=0

=>n=3(nhận)

Khi n=3 thì \(n^2+6n+10=3^2+6\cdot3+10=9+18+10=10+27=37\) là số nguyên tố

=>Nhận

TH2: \(n^2+6n+10=1\)

=>\(n^2+6n+9=0\)

=>\(\left(n+3\right)^2=0\)

=>n+3=0

=>n=-3(loại)

Vậy: n=3