Question 1197506
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Find the sum of the first 12 terms of 4,24,144,864,...
Geometric
{{{a[n] = a[1]*r^(n-1)}}}
  r = 6  n = 12
 {{{sum( a[i], i=1, n ) = a[1]((1-r^n)/(1-r))}}} 
        = {{{(4(1-6^12))/(1-6)}}}
Will let You finish  it Up
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