MTH 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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If a(n) = a(n-1) + b, with a(0) = 8, then if a( 260) = 0, what is the value of b?
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I am having a hard time with these type of problems. Here is what I did to work it out but I dont think this is the easiest way
a(260) = a(260-1) + b
= 259a + b = 0
a(0)=a(0-1)+b
a(0)=-1a + b = 8
(259)(-1a+b=8)
-259a + 259b = 2072
-1(259a + b = 0)
-259a + -1b = 0
subtracting one equation from the other now that a is eliminated to get
259b=2072
-(-1b=0)
260b=2072
260b/260=2072/260
b=518/65 or~7.969
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Is there an easier way, I remember doing these types of equations earlier and I kinda had a hard time with them then. I'm just trying to get prepared for the test. Thanks Billy
I believe you are interpreting the function notation of this situation as multiplication.
a(260), for example, is the value of the 260th member of the sequence. It does not stand for a * 260.
An alternative to representing the sequence as a(0), a(1), a(2), ... would be to use subscripts; however subscripts are not possible to represent as subscripts in a text file, where they would be written according to the form a_0, a_1, a_2, ...
So we choose to use the function notation.
Substituting first n = 1, then n = 2, then n = 3, etc., into the form a(n) = a(n-1) + b, we find that
a(n) = a(n-1) + b
means
a(1) = a(0) + b
a(2) = a(1) + b
a(3) = a(2) + b
etc..
a(1) = a(0) + b
a(2) = a(1) + b = (a(0) + b) + b = a(0) + 2 b. Using a(2) = a(0) + 2 b we get
a(3) = a(2) + b = a(0) + 2b + b = a(0) + 3 b. We could use a(3) = a(0) + 3 b to then get
a(4) = a(0) + 4 b, which we could use to get
a(5) = a(0) + 5 b, etc..
Extending this reasoning in the obvious manner, we conclude that
a(260) = a(0) + 260 b.
Now if a(0) = 8 and a(260) = 0, we get the equation
0 = 8 + 260 b,
which we easily solve for b, obtaining
b = -8 / 260.
Hopefully this clarifies the notation. If so, I recommend that you take another look at the relevant worksheet(s) and be sure you understand the related examples and problems. If not, be sure to submit additional questions.
MTH 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Find the first 4 terms of the sequence defined by a(n) = a(n-1) + -3/n, a(0) = -3.
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This is a similar equation like the one I sent you earlier. Thanks for your help.
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Is there an easier way, I remember doing these types of equations earlier and I kinda had a hard time with them then. I'm just trying to get prepared for the test. Thanks Billy
a(n) = a(n-1) - 3/n means that
a(1) = a(0) - 3/1
a(2) = a(1) - 3/2
a(3) = a(2) - 3/3
etc.
a(1) = a(0) - 3/1 = -3 - 3/1 = -3 - 3 = -6. So a(1) = -6.
a(2) = a(1) - 3/2 = -6 - 3/2 = -7/2. So a(2) = -7/2.
a(3) = a(2) - 3/3 = -7/2 - 3/3 = -9/2. SO a(3) = -9/2
etc..
Let me know if this, along with my response to your preceding question, don't clarify the notation and its application.