5 boys play a game of passing a ball from one to another. The boy 'A' starts the game. He throws the ball to another boy, who then throws the ball either back to 'A', or to a third boy, and so on.
Let a(n) be the number of ways that the ball returns to 'A' after n throws.
Explain why a(n)+a(n+1)=4^n.
Ball & Recurrence Sequence
2009-10-31 8:10 am
回答 (2)
2009-10-31 10:00 pm
✔ 最佳答案
Number of ways of getting the ball back to A after n throws is a(n).Number of ways of not getting the ball back to A after n throws
= Number of ways of getting the ball back to A after (n+1) throws
= a(n+1).
These two numbers must add up to the total number of ways of throwing balls n times which is equal to 4^n.
Hence, a(n) + a(n+1) = 4^n.
2009-10-31 9:31 am
This is trivial!
Number of ways of getting the ball back to A after n throws is a(n).
Number of ways of not getting the ball back to A after n throws
= Number of ways of getting the ball back to A after (n+1) throws
= a(n+1).
2009-10-31 01:31:46 補充：
These two numbers must add up to the total number of ways of throwing balls n times which is equal to 4^n.
Hence, a(n) + a(n+1) = 4^n.
Number of ways of getting the ball back to A after n throws is a(n).
Number of ways of not getting the ball back to A after n throws
= Number of ways of getting the ball back to A after (n+1) throws
= a(n+1).
2009-10-31 01:31:46 補充：
These two numbers must add up to the total number of ways of throwing balls n times which is equal to 4^n.
Hence, a(n) + a(n+1) = 4^n.
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