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?

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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

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