Solve the difference equation?

Question

Solve the difference equation y_n+1=sqrt((n+3)/(n+1))y_n in terms of the initial value y₀.

y₁=sqrt(3)y₀

y₂=sqrt(6)y₀

y₃=sqrt(10)y₀

Answer: y_n=y₀*sqrt(((n+2)(n+1))/2)

How do I get to the answer?

Answer 1

Substitute n for n+1,

y_n

= sqrt((n+2)/(n))y_n-1

= sqrt((n+2)(n+1)/(n(n-1)))y_n-2

= sqrt((n+2)(n+1)n/(n(n-1)(n-2)))y_n-3

= sqrt((n+2)(n+1)n...3/(n(n-1)(n-2)...1))y₀

= y₀*sqrt(((n+2)(n+1))/2), since (n+2)(n+1)n...3 = (n+2)!/2 = (n+2)(n+1)n!/2

 

QED