Expert: Paul Klarreich - 3/19/2010

Question
1 squared+3 squared+5 squared+... +(2n-1)squared=(4ncube-n)/3

Answer
Questioner: khalil
Country: Jamaica
Category: Advanced Math
Private: No
Subject: induction
Question: 1 squared+3 squared+5 squared+... +(2n-1)squared=(4ncube-n)/3


Your question is:

1^2 + 3^2 + ... + (2n-1)^2 = (4n^3 - n)/3

Induction proof:

1. STATE the theorem for n = 1.  Verify it is true.

2A. STATE the theorem for n = k.  Assume it is true.

2B. STATE the theorem for n = k+1.  Prove it is true. (use 2A)


1. 1^2 = (4(1)^3 - 1)/3
  1 = (4 - 1)/3
  1 = (3)/3
Yes.

2A. 1^2 + 3^2 + ... + (2k-1)^2 = (4k^3 - k)/3  <<< Assumed true.

2B. 1^2 + 3^2 + ... + (2(k+1)-1)^2 = (4(k+1)^3 - (k+1))/3  << To prove.

But :

1^2 + 3^2 + .............. + (2(k+1)-1)^2 =
1^2 + 3^2 + ... + (2k-1)^2 + (2(k+1)-1)^2 =
   (4k^3 - k)/3           + (2(k+1)-1)^2  (by assumption 2A)

Think you can do it from there?  It's just algebra.

If you get stuck, let me know.