We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>

www.vustudents.ning.com

 www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More

Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion

# CS702 Assignment No 01 Solution & Discussion Fall 2016

CS702 Assignment No 01 Solution & Discussion Fall 2016

+ How to Join Subject Study Groups & Get Helping Material?

+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?

+ VU Students Reserves The Right to Delete Your Profile, If?

Views: 1326

.

+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)

+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)

### Replies to This Discussion

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

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

Dear will you able to attach it in MSword form it seen some steps or words is missing...

2nd Question is easy to solve but in 1st question we have to convert 1st order Recurrence relation to 2nd order.

if u know then help... not getting it...

solve it for us

M.Tariq aap he help kr dain... jaldi

right solution for question no.2 of CS702 assignment?

question 1 plz

thanks :-) questions no 2 ka bsolution bata dain

## Latest Activity

7 minutes ago
Zari added a discussion to the group MTH301 Calculus II

### MTH301 Calculus II Assignment No 02 Fall 2019

21 minutes ago
M Ali replied to Hashim's discussion Eng301 Assignment 2 Solution 2020 & Fall 2019 in the group ENG301 Business Communication
30 minutes ago
33 minutes ago
34 minutes ago
1 hour ago
مخلص posted a discussion

### Samsung Galaxy Z Flip is going to change the GAME

1 hour ago
ZAINAB updated their profile
2 hours ago

### CS625 Assignment 3 Solution Jan 17-2020

2 hours ago
8 hours ago
+ ! ! ! ❣ maho ❣ ! + liked Rana Ali's discussion Galti ka mazaq
8 hours ago
8 hours ago

1

2

3