SOLUTION: Thanks for responding to my question. I understand what you were doing but I am still confuse about where the other equation works in f(n)= f(n-1)+n. Some other problems have thi
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-> SOLUTION: Thanks for responding to my question. I understand what you were doing but I am still confuse about where the other equation works in f(n)= f(n-1)+n. Some other problems have thi
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Question 67698This question is from textbook
: Thanks for responding to my question. I understand what you were doing but I am still confuse about where the other equation works in f(n)= f(n-1)+n. Some other problems have this equation changing. ie: f(n)= f(n-2) + f(n-1) and f(0)=1
Kate This question is from textbook
You can put this solution on YOUR website! I am still confuse about where the other equation works in f(n)= f(n-1)+n. Some other problems have this equation changing. ie: f(n)= f(n-2) + f(n-1) and f(0)=1
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f(n)=f(n-1)+n is the RULE for generating each term after the 1st term.
You were told f(1)=3 so you have the 1st term,
The next term is designated f(2).
The RULE tells you
f(2)=f(1)+2
f(2)=3+2
f(2)=5
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Now you can generate f(3)
The RULE tells you f(3)=f(3-1)+2 = f(2)+2
f(3)=5+2
f(3)=7
etc.
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Some RULES require you to use more than just the last term to generate
the next term. You have given an example of that in the following RULE:
f(n)= f(n-2) + f(n-1) and f(0)=1
The RULE require you to know not only the last term of the sequence;
you also must know the term before the last one.
f(n-1) is the last term you generated
f(n-2) is the term before that.
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If the RULE is f(n)= f(n-2) + f(n-1) and f(0)=1
you would also have to know f(1) in order to get started
f(2) would then be:
f(2) = f(0)+f(1)
You have f(0)=1 but you did not post f(1).
Maybe you have it but did not include it.
Email me if you want to further discuss these recursive problems.
Cheers,
Stan H.
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