Question 1178168
{{{a[1] }}} = {{{ (-2)^0/3^1}}}<br>

{{{a[2] }}} = {{{ (-2)^1/3^2}}}<br>

{{{a[3] }}} = {{{ (-2)^2/3^3}}}<br>

{{{a[4] }}} = {{{ (-2)^3/3^4}}}<br>

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{{{a[n]}}} = {{{ (-2)^(n-1)/3^n }}},  n=1,2,3,...