OK! Now Prove It
Congratulations to Phong Quach of Debney
Park Secondary College, Tomas from Malmesbury School, Sunil Ghosh
from the Royal Grammar School, Guildford, Chor Kiang Tan, Vassil
from Lawnswood High School, Tom from Madras College, Adam from King
James's School, Knaresborough, Alex from King Edward and Queen Mary
School, Lytham and Andaleeb from Woodhouse Sixth Form College,
London who all proved this result using mathematical induction.
Congratulations also to Daniel, and Alex, also to Mark and Eduardo
from the British School of Manila and to Michael and Sue of Madras
College and Yiwen of the Chines High School, Singapore for their
alternative method using the standard formulae .
First the method using the standard formulae: $$\sum_{i=1}^n i =
{1\over 2}n(n + 1)\quad {\rm and} \ \sum_{i=1}^n i^2 = {1\over
6}n(n + 1)(2n + 1).$$
We conjecture that $$\sum _{i=1}^n (2i - 1)^2 = {1 \over 6}(2n
-1)2n(2n+1).$$ Consider \begin{eqnarray} \\ \sum_{i=1}^n(2i - 1)^2
&=& 4 \sum_{i=1}^n i^2 - 4\sum_{i=1}^n i + \sum_{i=1}^n 1
\\ &=& 4\left(\frac{1}{6}n(n + 1)(2n + 1)\right ) -
4\left({1\over 2}n(n + 1)\right) + n \\ &=&
\frac{1}{6}(2n)(4n^2 + 6n + 2 - 6n - 6 + 3) \\ &=&
\frac{1}{6}(2n)(4n^2 - 1) \\ &=& \frac{1}{6}(2n)(2n - 1)(2n
+ 1) \end{eqnarray}
The second method uses mathematical induction.
The formulae given in the question are easily verified showing that
the conjecture is true for $n = 1, 2$ and $3$.
Suppose that the conjecture is true for $n = k$. Then $$1^2 + 3^2 +
... + (2k - 1)^2 = {(2k - 1)(2k)(2k + 1)\over 6}\quad (1)$$ Adding
one more term we get \begin{eqnarray} \\ 1^2 + 3^2 + ... + (2k - 1
)^2 + (2k + 1 )^2 &=& \frac{(2k - 1)(2k)(2k + 1)}{6} + (2k
+ 1)^2 \\ &=& \frac{(2k - 1)(2k)(2k + 1) + 6(2k + 1)^2}{6}
\\ &=& \frac{(2k + 1)(4k^2 + 10k + 6)}{6} \\ &=&
\frac{(2k + 1)(2k + 2)(2k + 3)}{6}. \end{eqnarray} This is
essentially the same as (1) but here $k$ is replaced by $k + 1$.
Thus if the conjecture is true for $n = k$, it is also true for $n
= k + 1$.
Thus by the axiom of induction $$1^2 + 3^2 + ... + (2n - 1)^2 =
{(2n - 1)(2n)(2n + 1)\over 6}.$$