assalamualaikum , tolong

[tex]a) p(2n) = \frac{3(2n)+8}{2n-1} = \frac{6n+8}{2n-1} [/tex]

[tex]b)p(n+1)-p(n)= \frac{3(n+1)+8}{(n+1)-1} - \frac{3n+8}{n-1} [/tex]
[tex]= \frac{3n+11}{n} - \frac{3n+8}{n-1} [/tex]
[tex]= \frac{(n-1)(3n+11)-n(3n+8)}{n(n-1)} [/tex]
[tex]= \frac{3n^{2}+8n-11-3n^{2}-8n}{n(n-1)} [/tex]
[tex]= \frac{-11}{n^{2}-n} [/tex]
[tex]=- \frac{11}{n^{2}-n}[/tex]