find the first four terms of the sequence given by the formula
d_n = 3n(n - 1)
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Work Shown:
Plug in n = 1 to find the first term
d_n = 3n(n-1)
d_1 = 3*1(1-1)
d_1 = 0
The first term is 0
Then repeat for n = 2
d_n = 3n(n-1)
d_2 = 3*2(2-1)
d_2 = 6
The second term is 6
We have enough to stop here since the sequence {0, 6, ...} is shown by choice D, while the other answer choices have a different two starting values.
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If you were to keep going, then it would look like this
Plug in n = 3
d_n = 3n(n-1)
d_3 = 3*3(3-1)
d_3 = 18
The third term is 18
and plug in n = 4
d_n = 3n(n-1)
d_4 = 3*4(4-1)
d_4 = 36
The fourth term is 36.
Answer:
d
Step-by-step explanation:
3n(n-1)
n = 1
3(1)(1-1) = 0
n = 2
3(2)(2-1) = 6
n = 3
3(3)(3-1) = 18
n = 4
3(4)(4-1) = 36