Question

Which of the following is a recursive formula for the sequence whose first four terms are 4,2,-4, and -22?
A. a^n = 3a_(n-1)-1
B. a^n=3a_(n-1)-10
C. a^n=-3a_(n-1)+1
D. a^n=-3a_(n-1)-10


2. Find the first six terms of the sequence in which a_1=4 and a_(n+1)+(n+6), for n≥1.
A. 4,11,19,28,38,49
B. 4,10,16,22,28,34
C. 4,12,21,31,42,54
D. 4,10,18,27,37,48

Answer 1

@whpalmer4

Answer 2

2= 3(4) -10
-4=3(2) -10
-22= 3(-4)-10
What is the answer?

Answer 3

This was for the first one

Answer 4

The second question is not written correctly

Answer 5

B. for the first one?

Answer 6

The second one is written: a_1=4 and a_n+1 = a_n + (n+6), for n≥1

Answer 7

for the first question is B the correct answer?

Answer 8

@eliassaab

Answer 9

B for the first one

Answer 10

4
11 = 4 +6+1
19= 11 + 7+1
28= 19 +8+1
39 =28 +9+1
50= 39 + 10+1
None of the choices are right

Answer 11

I got b, but i am not confident about my answer.

Answer 12

{4,11,19,28,39,50} is not one of the choices.
There is something wrong with the choices. Did you type them right:

Answer 13

4=a_1
a_2 = a_1 + 6+1=4+7 =11

Answer 14

The second term should be 11. I am going to gym

Answer 15

oh ok . thank you for your help :)