Question 1104282
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a(n) is the nth term while a(n-1) is the term just before the nth term. 
For example, if n = 5, then a(n) = a(5) is the fifth term while a(n-1) = a(5-1) = a(4) is the fourth term
Think of a(n) as function notation like f(x). Saying a(n) does NOT mean multiply.


Because this is a recursive sequence, we use each term to build the next term.
If we wanted to know the 10th term, then we need to find the first term, then the second, then the third all the way up to the 9th. 
We cannot simply find the 10th term by plugging in n = 10. 
This is a lot of work which is why recursive sequences are often a pain. 
Luckily we have a small number of terms to work with.


<font color=blue>a(1) = 4</font> is the given first term. We use this to find the second term
a(n) = 3*a(n-1) + 7
a(2) = 3*a(2-1) + 7 ... replace every n with 2
a(2) = 3*a(1) + 7
a(2) = 3*<font color=blue>a(1)</font> + 7 ... notice we have a(1)
a(2) = 3*<font color=blue>4</font> + 7 ... replace a(1) with 4
a(2) = 12+7
a(2) = 19
So the second term is 19


We repeat the same idea for the third term. Use the second term to find the third
a(n) = 3*a(n-1) + 7
a(3) = 3*a(3-1) + 7 ... replace every n with 3
a(3) = 3*a(2) + 7 ... notice we have a(2)
a(3) = 3*19 + 7 ... replace that a(2) with 19
a(3) = 57 + 7
a(3) = 64
The third term is 64


And again for the fourth
a(n) = 3*a(n-1) + 7
a(4) = 3*a(4-1) + 7 ... replace every n with 4
a(4) = 3*a(3) + 7
a(4) = 3*64 + 7 ... replace a(3) with 64
a(4) = 192 + 7
a(4) = 199
The fourth term is 199


Then finally the fifth
a(n) = 3*a(n-1) + 7
a(5) = 3*a(5-1) + 7 ... replace every n with 5
a(5) = 3*a(4) + 7
a(5) = 3*199 + 7 ... replace a(4) with 199
a(5) = 597 + 7
a(5) = 604
The fifth term is 604


So the first five terms are: <font size=4 color=red>4, 19, 64, 199, and 604</font>
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