Find the first four terms of the sequence given by the formula
d_n = 3n(n - 1)

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.

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.

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

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says