Transcript Inductive Step

Problems to Solve Involving
Induction
n
 ar
k

ar
n 1
a
r 1
k 0
,r  1 ?
Proof by Induction
Basis Step: Does it work for n=0?
0
 ar
k
0
 ar  a
k 0
ar
0 1
a
r 1

ar  a
r 1

a(r  1)
r 1
 a
Inductive Step
Assume that for r1
n
 ar
k

ar
k 0
n 1
a
r 1
We must show that
n 1
 ar
k 0
k

ar
( n1 ) 1
r 1
a
n 1
 ar
k
 ar
n 1
n

k 0



ar
 ar
k
 ar
k 0
n 1
(r  1 )  ar
n 1
a
r 1
ar
n 2
 ar
n1
 ar
r 1
ar
( n1 ) 1
r 1
a
n1
a
n 1

ar
n1
a
r 1
n
Prove that  (3 j  2 ) is
2
O(n )
j 1
First let’s find a closed form solution for the
sum.
n
n
n
j 1
j1
j1
 (3 j  2 )  3  j  2  1
= 3n(n+1)/2 - 2n = (3n2 + 3n -4n)/2 = (3n2 -n)/2
How can we be sure that our
arithmetic is correct?
n
Prove that
 (3 j  2 ) 
2
3n  n
2
j 1
Proof by Induction
Basis Step: 1
For n=1
 (3 j  2 )  3(1)  2  1
j 1
3(1) 2  1
2

2
2
1
Inductive Step
n
Assume that
 (3 j  2 ) 
2
3n  n
2
j 1
Then we must show that:
n1
 (3 j  2 ) 
j 1
2
3( n  1)  ( n  1)
2

n1

n
 (3 j  2 )   (3 j  2)   3( n  1)  2
 j  1
j 1

3n 2  n
2

 3(n  1)  2
2


3n  n  6( n  1)  4
2
3n 2  5n  2
2

3n 2  6n  3  n  1
3(n 2  2 n  1)  ( n  1)
2
2

3( n  1) 2  ( n  1)
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 n1?
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
n
2
n
2
Prove that every by
(n > 1)
chessboard can be tiled with T-ominos
Basis Case:
2
2
x
2
2 Board
Basis Case:
2
2
x
2
2 Board
Basis Case:
2
2
x
2
2 Board
Basis Case:
2
2
x
2
2 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.