Question 380241
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Hi,
{{{a[n] = 4n!/(n+3)}}}
{{{a[1] = 4*1!/(1+3)= 1}}}
{{{a[2] = 4*2!/(2+3)= 8/5}}}
{{{a[3] = 4*3!/(3+3)= 24/6 = 4}}}
{{{a[4] = 4*4!/(4+3)= 96/7}}}
{{{a[5] =4*5!/(5+3)= 384/8 = 48}}}