Hey Lindskou.

I figure you are having trouble, so I'll start you with a hint.

For the first one you have y(n+1) = a(n)y(n) + c(n). Now going along you have

y(n+2)

= a(n+1)y(n+1) + c(n+1)

= a(n+1)[a(n)y(n) + c(n)] + c(n+1)

= a(n+1)a(n)y(n) + c(n)a(n+1) + c(n+1)

y(n+3)

= a(n+3)y(n+3) + c(n+3)

= a(n+3)[y(n+2)a(n+2) + c(n+2)] + c(n+3)

= a(n+3)a(n+2)a(n+1)a(n)y(n) + c(n)a(n+3)a(n+2)a(n+1) + a(n+3)a(n+2)c(n+1) + a(n+3)c(n+2) + c(n+3)

Now if the a's and c's are constant you got in the above

= a^3y(n) + ca^3 + ca^2 + ac + c

= a^3y(n) + c[a^3 + a^2 + a + 1]

which is approaching that formula.

For back-substitution you should use y(n) (from the series) with y(n+1) (use n = n+1 instead of a normal n) and then see if you can re-arrange the expression so that you get the solution for y(n+1) = a(n)y(n) + c(n) and this is simply expanding the series out on both sides and collecting terms.