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CS702 Assignment No 01 Solution & Discussion Fall 2016
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Cs702
Idea Solution
Prove the following by induction.
Solution:
We need to identify the three basic elements of the proof.
Induction Hypothesis: It is what we need to prove,
Base Case: We need to show that the induction hypothesis holds for the value of n = 1.
This is easy to show.
The left hand side of the equality is
When n = 1, the value of the left hand side is
The right hand side in this case is equal to
Therefore, for the base case, the left hand side is equal in value to the
right hand side. In other words, the induction hypothesis holds for the base case.
Induction Step: Here we need to show that if the induction hypothesis holds for an arbitrary positive integer value k (and for each positive integer smaller than k), it holds for the next higher integer k + 1. Let us assume that for some k >= 1, the induction hypothesis holds.
That is,
We need to show that
.
The proof follows. Let us start from the right hand side of the equation given immediately
above.
The sequence<un>is defined by the recurrence
You have initial conditionu1 = 1:
Now you have to showunin terms of Fibonacci / Lucasnumbers.
We first prove a straightforward lemma.
Lemma: For all n>=1;
(1), (2)
Proof (of the Lemma): From straightforward manipulations of the Binet formulas for Fn and Ln; one can easily show that
(3)
and
(4)
Thus we have
From (3)
Which is (1) also,
From (4)
Which is (2) also
Now we can prove the result in Question.
Theorem: For all n>=1,
Let so our goal is to prove that for all n>=1 it is clear that
Hence we need only check that unsatisfies
Notice that
From Lemma above
2nd Question is easy to solve but in 1st question we have to convert 1st order Recurrence relation to 2nd order.
right solution for question no.2 of CS702 assignment?
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