Is this correct?

  • Thread starter Chromium
  • Start date
  • #1
56
0
P(n): 1 + 3 + 5 + ... + (2n-1) = n^2

Prove P(1)
P(1) = 1 = 1^2
P(1) is true

(A) P(k): 1 + 3 + 5 +...+ (2k-1) = k^2

(B) P(k + 1): 1 + 3 + 5 +...+ (2k-1) + (2(k+1)-1)
or
(2k-1) + (2k + 1) = (k+1)^2

Assuming A, prove B

(2k-1) = k^2
(2(k+1)-1) = (k+1)(k+1)
(2k + 1) = (k^2 + 2k+ 2)
= (k+1)^2

When in comes to the inductive step, does it "differ" from problem to problem? I can always get to the inductive assumption, but then I'm never sure just how to go about proving it.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
961
P(n): 1 + 3 + 5 + ... + (2n-1) = n^2

Prove P(1)
P(1) = 1 = 1^2
P(1) is true

(A) P(k): 1 + 3 + 5 +...+ (2k-1) = k^2

(B) P(k + 1): 1 + 3 + 5 +...+ (2k-1) + (2(k+1)-1)
or
(2k-1) + (2k + 1) = (k+1)^2
I'm, sorry? how does 1+ 3+ ...+ 2(k+1)= (2k-1)+ 2k???

Assuming A, prove B

(2k-1) = k^2
(2(k+1)-1) = (k+1)(k+1)
(2k + 1) = (k^2 + 2k+ 2)
= (k+1)^2

When in comes to the inductive step, does it "differ" from problem to problem? I can always get to the inductive assumption, but then I'm never sure just how to go about proving it.
You want to prove that "P(k)= (k+1)^2. P(k+1)= P(k)+ 2k+1 for every k. The "induction hypothesis" is that P(k-1)= k^2 . From that, P(k+1)= P(k)+ k+1= k^2+ 2k+ 1= (k+1)^2/
 

Related Threads on Is this correct?

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
21
Views
2K
Replies
2
Views
825
  • Last Post
Replies
5
Views
1K
Top