Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $ Prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $.  I provide the answer below.  
 A: We prove that $ 1^2 + 3^2 + ... + (2n-1)^2 = \displaystyle \frac{4n^3 -n}{3} $. 
Base Case:  Let $ n=1 $.  Then 
$  1^2 + 3^2 + ... + (2n-1)^2 = (2*1-1)^2 = 1 =  \displaystyle \frac{4*1^3 - 1}{3} = \frac{4n^3 -n}{3} $.  
Inductive Step:  Assume that $  1^2 + 3^2 + ... + (2k-1)^2 = \displaystyle \frac{4k^3 -k}{3} $ for some   $ k \in \textbf{N} $. 
We show that $  1^2 + 3^2 + ... + (2k-1)^2 + (2(k+1)-1)^2 = \displaystyle \frac{4(k+1)^3-(k+1)}{3} $.  
Due to the assumption, it is sufficient to show that 
$ \displaystyle \frac{4k^3-k}{3} +  (2(k+1)-1)^2 = \displaystyle  \frac{4(k+1)^3-(k+1)}{3} $.
Note the following expansions
$ 4(k+1)^3 - (k+1) = 4k^3 + 12k^2 + 12k + 4 - k - 1 = 4k^3 + 12k^2 + 11k + 3  $ 
and
$ (2(k+1)-1)^2 = (2k+1)^2 = 4k^2 +4k + 1 $.   
So $ \displaystyle \frac{4k^3-k}{3} +  (2(k+1)-1)^2 = \frac{4k^3 - k + 12k^2 + 12k + 3}{3} = \frac{4k^3 + 12k^2 + 11k + 3}{3} $  
$ =  \displaystyle \frac{4(k+1)^3-(k+1)}{3} $. 
A: The tag says "induction," so step one: prove that
$$\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$$
by induction.
Step two: Given that
$$f(n)=\sum_{i=1}^n i^2=\frac {n(n+1)(2n+1)}6$$
we have
$$f(2n-1)=\frac {2n(2n-1)(4n-1)}6=\frac {n(2n-1)(4n-1)}3$$
$$4f(n-1)=2\frac {n(n-1)(2n-1)}3=4\sum_{i=1}^{n-1} i^2=\sum_{i=1}^{n-1} (2i)^2$$
and
$$f(2n-1)-4f(n-1)=\sum_{i=1}^n(2i-1)^2=\frac{n(2n-1)(4n-1)}3-2\frac{n(n-1)(2n-1)}3$$
$$=\frac{(n(4n-1)-2n(n-1))(2n-1)}3=\frac {(4n^2-n-2n^2+2n)(2n-1)}3$$
$$=\frac{(2n^2+n)(2n-1)}3=\frac{4n^3-n}3$$
