Solve the difference equation y_{n+1}=sqrt((n+3)/(n+1))y_{n} in terms of the initial value y_{0}.
y_{1}=sqrt(3)y_{0}
y_{2}=sqrt(6)y_{0}
y_{3}=sqrt(10)y_{0}
Answer: y_{n}=y_{0}*sqrt(((n+2)(n+1))/2)
How do I get to the answer?
Solve the difference equation y_{n+1}=sqrt((n+3)/(n+1))y_{n} in terms of the initial value y_{0}.
y_{1}=sqrt(3)y_{0}
y_{2}=sqrt(6)y_{0}
y_{3}=sqrt(10)y_{0}
Answer: y_{n}=y_{0}*sqrt(((n+2)(n+1))/2)
How do I get to the answer?
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))y_{0}
= y_{0}*sqrt(((n+2)(n+1))/2), since (n+2)(n+1)n...3 = (n+2)!/2 = (n+2)(n+1)n!/2
QED