Jika p (n)=(n+2)(n+3) per (n+1)(n+3) tentukanlah p=(n+1)

p(n)=(n+2)(n+3) per (n+1)(n+3)
p(n+1)=(n+1+2)(n+1+3) per (n+1+1)(n+1+3)
=(n+3)(n+4) per (n+2)(n+4)

[tex]p(n)=\frac{(n+2)(n+3)}{(n+1)(n+3)}\\p(n+1)=\frac{((n+1)+2)((n+1)+3)}{((n+1)+1)((n+3)+1)}\\p(n+1)=\frac{(n+3)(n+4)}{(n+2)(n+4)}[/tex]