Given:
n/2 if n is even
3n + 1 if n is odd
To find: The value
Solution:
f(n)= { n/2 if n is even
f(2)= 2/2
=1
f(4)= 4/2
= 2
f(6)= 6/2
=3
In every consecutive even integer the common sequence is +1
f(n)= 3n + 1
f(1) = 3(1)+1
=3 + 1
= 4
f(3)= 3(3) + 1
= 9 + 1
= 10
f(5)= 3(5) + 1
= 15 + 1
= 16
In every consecutive odd integer the common sequence is +6
Answer:
f(n) = {1,2,3,4,5,6...}
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Answers & Comments
Given:
n/2 if n is even
3n + 1 if n is odd
To find: The value
Solution:
f(n)= { n/2 if n is even
f(2)= 2/2
=1
f(4)= 4/2
= 2
f(6)= 6/2
=3
In every consecutive even integer the common sequence is +1
f(n)= 3n + 1
f(1) = 3(1)+1
=3 + 1
= 4
f(3)= 3(3) + 1
= 9 + 1
= 10
f(5)= 3(5) + 1
= 15 + 1
= 16
In every consecutive odd integer the common sequence is +6
Answer:
f(n) = {1,2,3,4,5,6...}