Przekształcając wzór ciągu a_n mamy:

$a_n=\left(p+1\right)\cdot \frac{{\left(n+5\right)}^2}{\left(n+1\right)\left(n+2\right)}+\left(2p+2\right)\cdot \frac{{\left(n+5\right)}^2}{\left(n+2\right)\left(n+3\right)}$              

 

Zauważmy, że:

$\underset{n\to \infty }{\lim }\left(p+1\right)\frac{{\left(n+5\right)}^2}{\left(n+1\right)\left(n+2\right)}=\underset{n\to \infty }{\lim }\left(p+1\right)\frac{n^2+10n+25}{n^2+2n+n+2}=$  

$=\underset{n\to \infty }{\lim }\left(p+1\right)\frac{n^2+10n+25}{n^2+3n+2}=\underset{n\to \infty }{\lim }\left(p+1\right)\frac{n^2\left(1+\frac{10}{n}+\frac{25}{n^2}\right)}{n^2\left(1+\frac{3}{n}+\frac{2}{n^2}\right)}=$  

$=\underset{n\to \infty }{\lim }\left(p+1\right)\frac{1+\frac{10}{n}+\frac{25}{n^2}}{1+\frac{3}{n}+\frac{2}{n^2}}=p+1$  

 

$\underset{n\to \infty }{\lim }\left(2p+2\right)\cdot \frac{{\left(n+5\right)}^2}{\left(n+2\right)\left(n+3\right)}=\underset{n\to \infty }{\lim }\left(2p+2\right)\cdot \frac{n^2+10n+25}{n^2+3n+2n+6}=$  

$=\underset{n\to \infty }{\lim }\left(2p+2\right)\cdot \frac{n^2+10n+25}{n^2+5n+6}=\underset{n\to \infty }{\lim }\left(2p+2\right)\cdot \frac{n^2\left(1+\frac{10}{n}+\frac{25}{n^2}\right)}{n^2\left(1+\frac{5}{n}+\frac{6}{n^2}\right)}=$  

$=\underset{n\to \infty }{\lim }\left(2p+2\right)\cdot \frac{1+\frac{10}{n}+\frac{25}{n^2}}{1+\frac{5}{n}+\frac{6}{n^2}}=2p+2$  

 

Zatem otrzymujemy:

$\underset{n\to \infty }{\lim }{a}_n=p+1+2p+2=3p+3$  

Wiemy, że granica ciągu jest równa 12, zatem:

$3p+3=12\mid :3$  

$p+1=4\mid -1$  

$p=3$