Problems to Solve Involving Induction

Proof by Induction

Basis Step: Does it work for n=0?

ark

k0

n

arn1 a

r 1,r 1?

ark

k0

0

ar0 a

ar01 ar 1

ar ar 1

a(r 1)

r 1a

Inductive Step

Assume that for r1

We must show that

ark

k0

n

arn1 a

r 1

ark

k0

n1

ar(n1)1 a

r 1

ark

k0

n1

arn1 arkk0

n

arn1 arn1 ar 1

arn1(r 1) arn1 a

r 1

arn2 arn1 arn1 a

r 1

ar(n1)1 ar 1

Prove that is O(n2)

(3j 2j1

n

)

First let’s find a closed form solution for the sum.

= 3n(n+1)/2 - 2n = (3n2 + 3n -4n)/2 = (3n2 -n)/2

(3j 2

j1

n

) 3 jj1

n

2 1j1

n

How can we be sure that our arithmetic is correct?

Prove that

Proof by Induction

Basis Step:

For n=1

(3j 2

j1

n

) 3n2 n

2

(3 j 2j1

1

) 3(1) 2 1

3(1)2 1

2

2

21

Inductive Step

Assume that

Then we must show that:

(3j 2

j1

n

) 3n2 n

2

(3j 2

j1

n1

) 3(n1)2 (n1)

2

(3j 2j1

n1

) (3 j 2)j1

n

3(n1) 2

3n2 n2

3(n1) 2

3n2 n 6(n1) 4

2

3n2 5n 2

2

3n2 6n 3 n 1

2

3(n2 2n1) (n1)

2

3(n 1)2 (n1)

2

Big-Oh

(3n2 -n)/2 is clearly O(n2) since

(3n2 -n)/2 (3n2 -n) 3n2 since n is positive.

So chose C = 3 and k=1 and we have

| (3n2 -n)/2| (3) |n2 | when n > 1.

Does 1*1! + 2*2! + 3*3!+…+n*n! = (n+1)! -1 n1?

Basis Step: n = 1

1*1! = 1, (1+1)!-1 = 2-1=1

Inductive Step:

Assume that

1*1! + 2*2! + 3*3!+…+n*n! = (n+1)! -1

We must show that

1*1!+2*2!+3*3!+…+n*n!+(n+1)*(n+1)! =(n+2)!-1

1*1!+2*2!+3*3!+…+n*n!+(n+1)*(n+1)!

= (n+1)! -1 + (n+1)*(n+1)!

= (n+1+1)(n+1)! -1

= (n+2)(n+1)! - 1

= (n+2)! - 1

Prove that every 2n by 2n (n > 1) chessboard can be tiled with T-ominos

Basis Case: 22 x 22 Board

Basis Case: 22 x 22 Board

Basis Case: 22 x 22 Board

Basis Case: 22 x 22 Board

Inductive Step

Assume that we can tile a board of size 2n by 2n. We must show that this implies that we can tile a board of size 2n+1 by 2n+1.

Proof: Divide the 2n+1 by 2n+1 board into 4 parts, each of size 2n by 2n. Since we know that each of these boards can be tiled, then we can put them together to tile the 2n+1 by 2n+1 board.