# Write the first five terms of each of the sequences. whose nth terms are an = n(n2 + 5)/4

an = n(n2 + 5)/4

Substituting n = 1, 2, 3, 4 and 5 we get

a1 = 1(12 + 5)/4 = 6/4 = 3/2

a2 = 2(22 + 5)/4 = 2(4 + 5)/4 = 18/4 = 9/2

a3 = 3(32 + 5)/4 = 3(9 + 5)/4 = 135/2

a4 = 4(42 + 5)/4 = 4(16 + 5)/4 = 84/4 = 21

a5 = 5(52 + 5)/4 = 5(25 + 5)/4 = 150/2 = 75

Therefore the first term is 3/2, 9/2, 135/2, 21 and 75/2.

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