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(9n2 - 121) = (3n + 11) (3n - 11)
3n(n+1] + 5 is the nth term
5
The Series has the formula 3n + 1/2(n - 1)(n - 2) = 3n + 1/2(n2 - 3n + 2) which simplifies to, 1/2(n2 +3n + 2)
3n and n = 4? 3n ( the four replaces the n 3(4) = 12 =======