<s>[INST] 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.  
 [/INST] 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
$$f(2n-1)-4f(n-1)=\sum_{i=1}^n(2i-1)^2=\frac{n(2n-1)(4n-1)}3-2\frac{n(n-1)(2n-1)}3$$
$$=\frac{(n(4n-1)-2n(n-1))(2n-1)}3=\frac {(4n^2-n-2n^2+2n)(2n-1)}3$$
$$=\frac{(2n^2+n)(2n-1)}3=\frac{4n^3-n}3$$
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