the nth term of a sequence is given by an²+bn where a and b are integers the 2nd term of the sequence is 2 4th term of the sequence is 20. find the 6th term of the sequence

54

Step-by-step explanation:

Tn = an²+bn

T2 = a2²+2b

2 = 4a +2b

Multiply all through by 2

4 = 8a +4b

4b = 4-8a

T4 = a4² + 4b

20 = 16a +4b

4b = 20-16a

4-8a = 20-16a

16a -8a = 20-4

8a = 16

a = 16/8 = 2

4b = 4-8a = 4-8(2) = 4-16 = -12

b = -12/4 = -3

T6 = a6² + 6b

T6 = 2(36) + 6(-3) = 72 -18 = 54

54

Step-by-step explanation:

c

step-by-step explanation:

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~