Answer :
Problem 4a
The instructions are incomplete. You set up the recursive formula, but didn't ask any question about said formula.
I'll assume that your teacher wants you to list out a few terms. I'll list out the first five terms.
The notation a_1 = 4 is the same as writing [tex]a_1 = 4[/tex] where the '1' is a subscript. It tells us that the first term is 4.
The nth term a_n or [tex]a_n[/tex] is defined as such
[tex]a_n = 2*(1/3 + a_{n-1})\\\\[/tex]
Notice how if we replaced n with 2, then we get
[tex]a_n = 2*(1/3 + a_{n-1})\\\\a_2 = 2*(1/3 + a_{2-1})\\\\a_2 = 2*(1/3 + a_1)\\\\[/tex]
So the second term is directly tied to the first term, or it is dependent on it.
We'll replace a_1 with 4 to get the following
[tex]a_2 = 2*(1/3 + a_1)\\\\a_2 = 2*(1/3 + 4)\\\\a_2 = 2*(1/3 + 12/3)\\\\a_2 = 2*(13/3)\\\\a_2 = 26/3\\\\[/tex]
So the second term is 26/3.
As you can guess, the third term is going to be found in a similar fashion
[tex]a_n = 2*(1/3 + a_{n-1})\\\\a_3 = 2*(1/3 + a_{3-1})\\\\a_3 = 2*(1/3 + a_2)\\\\a_3 = 2*(1/3 + 26/3)\\\\a_3 = 2*(27/3)\\\\a_3 = 2*(9)\\\\a_3 = 18\\\\[/tex]
So 18 is the third term.
We'll repeat for n = 4 to get the fourth term.
[tex]a_n = 2*(1/3 + a_{n-1})\\\\a_4 = 2*(1/3 + a_{4-1})\\\\a_4 = 2*(1/3 + a_3)\\\\a_4 = 2*(1/3 + 18)\\\\a_4 = 2*(1/3 + 54/3)\\\\a_4 = 2*(55/3)\\\\a_4 = 110/3\\\\[/tex]
The fourth term is 110/3.
Lastly, we'll plug in n = 5
[tex]a_n = 2*(1/3 + a_{n-1})\\\\a_5 = 2*(1/3 + a_{5-1})\\\\a_5 = 2*(1/3 + a_4)\\\\a_5 = 2*(1/3 + 110/3)\\\\a_5 = 2*(111/3)\\\\a_5 = 2*(37)\\\\a_5 = 74\\\\[/tex]
The fifth term is 74.
Answer: The first five terms are 4, 26/3, 18, 110/3, 74
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Problem 4b
Again, the instructions are missing. I'll assume the same thing as problem 4a.
[tex]a_1 = 6[/tex] is the first term
Plug n = 2 into the first equation to get
[tex]a_n = \frac{n}{a_{n-1}}\\\\a_2 = \frac{2}{a_{2-1}}\\\\a_2 = \frac{2}{a_{1}}\\\\a_2 = \frac{2}{6}\\\\a_2 = \frac{1}{3}\\\\[/tex]
The second term is 1/3.
Repeat for n = 3
[tex]a_n = \frac{n}{a_{n-1}}\\\\a_3 = \frac{3}{a_{3-1}}\\\\a_3 = \frac{3}{a_{2}}\\\\a_3 = \frac{3}{1/3}\\\\a_3 = 3\div\frac{1}{3}\\\\a_3 = 3\times\frac{3}{1}\\\\a_3 = 9\\\\[/tex]
The third term is 9
Repeat for n = 4.
[tex]a_n = \frac{n}{a_{n-1}}\\\\a_4 = \frac{4}{a_{4-1}}\\\\a_4 = \frac{4}{a_{3}}\\\\a_4 = \frac{4}{9}\\\\[/tex]
The fourth term is 4/9
Repeat for n = 5
[tex]a_n = \frac{n}{a_{n-1}}\\\\a_5 = \frac{5}{a_{5-1}}\\\\a_5 = \frac{5}{a_{4}}\\\\a_5 = 5 \div a_{4}\\\\a_5 = 5 \div \frac{4}{9}\\\\a_5 = 5 \times \frac{9}{4}\\\\a_5 = \frac{5}{1} \times \frac{9}{4}\\\\a_5 = \frac{5*9}{1*4}\\\\a_5 = \frac{45}{4}\\\\[/tex]
Answer: The first five terms are 6, 1/3, 9, 4/9, 45/4
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Problem 4c
I'm not much help here for this problem. Not only are the instructions missing, but it's not clear how this sequence is set up. If I had to guess, it's somehow recursively defined. How exactly, I'm not sure. I would ask your teacher for any clarification.