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Consider Tn = 4n³- 6n²+ Tn-1 where T0 = 2, then

A) T1=0, T2=8, T3=62
B) T1=2, T2=8, T3=62
C) T1=2, T2=8, T3=68
D) T1=0, T2=8, T3=68

1 Answer

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Final answer:

The first three terms of the sequence defined by Tn = 4n³ - 6n² + Tn-1 with T0 = 2 are calculated as T1=0, T2=8, and T3=62, corresponding to option A.

Step-by-step explanation:

The question asks to calculate the first three terms of the sequence defined by the recurrence relation Tn = 4n³ - 6n² + Tn-1 with the initial condition T0 = 2.

To find T1, plug n=1 into the equation:

  • T1 = 4(1)³ - 6(1)² + T0
  • T1 = 4 - 6 + 2
  • T1 = 0

Now to find T2, substitute n=2:

  • T2 = 4(2)³ - 6(2)² + T1
  • T2 = 32 - 24 + 0
  • T2 = 8

Finally, we calculate T3:

  • T3 = 4(3)³ - 6(3)² + T2
  • T3 = 108 - 54 + 8
  • T3 = 62

Thus, the correct answer is T1=0, T2=8, T3=62, which corresponds to option A).

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