Question

solve recurrence relation S(k)-4S(k-1)+3S(k-2)= 5k
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Answer 1

Answer:

COMBINATORICS

 

1 MATHEMATICAL INDUCTION

2 STRONG INDUCTION & WELL ORDERING

3 RECURRENCE RELATION

4 SOLVING LINEAR RECURRENCE RELATIONS

5 GENERATING FUNCTION

6 THE PRINCIPLE OF INCLUSION & EXCLUSION

 

 

Pigeon Hole Principle:

 

If (n=1) pigeon occupies ‘n’ holes then atleast one hole has more than 1 pigeon. 

 

Proof:

 

Assume (n+1)   pigeon   occupies   „n‟   holes.

 

Claim: Atleast one hole has more than one pigeon.

 

Suppose not, ie. Atleast one hole has not more than one pigeon. 

Therefore, each and every hole has exactly one pigeon.

Since, there are ‘n’ holes, which implies, we have totally ‘n’ pigeon. 

 

Which is a Þ Ü to our assumption that there are (n+1) pigeon. 

Therefore, atleast one hole has more than 1 pigeon.

Answer 2

Answer:

To solve recurrence relation S(k)-4S(k-1)+3S(k-2)= 5k

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Step-by-step explanation:

Given that: solve recurrence relation S(k)-4S(k-1)+3S(k-2)= 5k

Here we are provided with right hand side(R.H.S)= S(k)-4S(k-1)+3S(k-2)

And left hand side(L.H.S)= 5k

We have to solve R.H.S and L.H.S to show the relation between them.

Therefore first taking R.H.S S(k)-4S(k-1)+3S(k-2)

SK-4SK+4S+3SK-6S

-3SK+3SK+4S-6S

-5S...…(on solving R.H.S)

Now equating L.H.S and R.H.S

We get,

-5S=5K

-S=K

Hence we get. -S=K

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