Show that Pn satisfy the following recursion Pn+1 = P&Pn-k- k=0

College Algebra (MindTap Course List)
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Author:R. David Gustafson, Jeff Hughes
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Chapter8: Sequences, Series, And Probability
Section8.CR: Chapter Review
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Let us consider parenthesis, as used in a sentence for the written English
language. The only rule for a single parenthesis is that it must open, we write " (" for that, and it
must close, in which case we write ")".
If we use a parenthesis inside another parenthesis, we must make sure that we close the parenthesis
inside first, before closing the external parenthesis. Otherwise it is not correctly written. For instance,
0) is correct, but )(() is not correct. Let Pn be the number of ways in which n parenthesis can be
written correctly. The start of the sequence is
(Ро, Р, Ра, Рз, ...) %3D (1, 1, 2, 5,...),
corresponding to no parenthesis, which gives Po = 1, the unique parenthesis (), which gives P = 1,
P2 = 2 because we can write ()) and (()) and finally there are five ways, P3 = 5, of correctly writing
MAT 108: PROBLEM SET 3
3
three parenthesis:
)00,0(0),(0)0), (O0),((O)).
Show that Pn satisfy the following recursion
n
Pn+1 => Pk Pn-k-
k=0
Transcribed Image Text:Let us consider parenthesis, as used in a sentence for the written English language. The only rule for a single parenthesis is that it must open, we write " (" for that, and it must close, in which case we write ")". If we use a parenthesis inside another parenthesis, we must make sure that we close the parenthesis inside first, before closing the external parenthesis. Otherwise it is not correctly written. For instance, 0) is correct, but )(() is not correct. Let Pn be the number of ways in which n parenthesis can be written correctly. The start of the sequence is (Ро, Р, Ра, Рз, ...) %3D (1, 1, 2, 5,...), corresponding to no parenthesis, which gives Po = 1, the unique parenthesis (), which gives P = 1, P2 = 2 because we can write ()) and (()) and finally there are five ways, P3 = 5, of correctly writing MAT 108: PROBLEM SET 3 3 three parenthesis: )00,0(0),(0)0), (O0),((O)). Show that Pn satisfy the following recursion n Pn+1 => Pk Pn-k- k=0
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