Question 975258
{{{A[n]=(3^(n+2)/2^n)}}}
{{{A[1]=(3^(1+2)/2^1)=3^3/2^1=27/2}}}
{{{A[2]=(3^(2+2)/2^2)=3^4/2^2=81/4}}}
{{{A[3]=(3^(3+2)/2^3)=3^5/2^3=243/8}}}
{{{A[4]=(3^(4+2)/2^4)=3^6/2^4=729/16}}}
or
{{{A[n]=(3^(n)+2)/2^n)}}}
{{{A[1]=(3^(1)+2)/2^1=(3+2)/2^1=5/2}}}
{{{A[2]=(3^(2)+2)/2^2=(9+2)/2^2=11/4}}}
{{{A[3]=(3^(3)+2)/2^3=(27+2)/2^3=29/8}}}
{{{A[4]=(3^(4)+2)/2^4=(81+2)/2^4=83/16}}}