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Question 1104282: A sequence is defined recursively by the given formulas. Find the first five terms of the sequence.
an = 3a(n − 1) + 7 and a1 = 4
I thought it would be
a1=4
a2= 3(2-1)+7=10
a3= 3(3-1)+7=13
a4= 3(4-1)+7=16
a5= 3(5-1)+7=19
But those are wrong. What am I missing with this?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
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.
a(1) = 4 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*a(1) + 7 ... notice we have a(1)
a(2) = 3*4 + 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: 4, 19, 64, 199, and 604
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