Answers to Practice Test 4
Proofs by Induction. Use your blue books.
n
8. › i(i!) = (n+1)! - 1.
i=1
Base Case. Let n = 1. It suffices to prove that 1 ∑ 1! = (1 + 1)! - 1
1 ∑ 1! = 1 ∑ 1
= 1
= 2 - 1
= 2! - 1
= (1 + 1)! - 1
Inductive Case.
Inductive hypothesis: the sum from 1 to k = (k + 1)! - 1
It suffices to show that the sum from 1 to k + 1 = (k + 2)! - 1.
Sum from 1 to k + 1 = sum from 1 to k + (k+1)∑(k+1)!
= (k + 1)! - 1 + (k+1)∑(k+1)!, by IH.
= (k + 2) ∑ (k+1)! - 1
9. Consider the formal language consisting of vocabulary items
k
and
=
. Grammatical
strings are defined recursively as follows:
(1) List(
k
) is a string.
(2) If è is a string, then List(
k
)^è^List(
=
) is a string.
(3) There are no other strings.
Prove that in any string, the number of
k
's is greater than the number of
=
's.
First, we use recursionn to define two functions, the number of k’s function (n) and the
number of =’s function (q):
n(List(k)) = 1
n(List(k)^è^List(=)) = n(è) + 1
q(List(k)) = 0
q(List(k)^è^List(=)) = q(è) + 1
Let ç be an arbitrary string
Base Case. String ç = List(k).
n(ç) = 1, q(ç) = 0, and 1 > 0.
Inductive case. ç = List(k)^ß^List(=).
Inductive hypothesis: n(ß) > q(ß).
n(ç) = n(ß) + 1, and q(ç) = q(ß) + 1,
So n(ç) > q(ç).
