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

Question 1149333: 1^2+3^2+...+(2n-1)^2=1/3n(4n^2-1)
Found 2 solutions by ikleyn, math_helper:
Answer by ikleyn(52867) (Show Source): You can put this solution on YOUR website!
.

There are several proofs by the method of Mathematical induction.

See the links

https://socratic.org/questions/show-by-induction-that-aa-n-1-1-2-3-2-5-2-2n-1-2-n-3-4n-2-1

https://math.stackexchange.com/questions/623504/prove-that-12-32-2n-12-displaystyle-frac4n3-n3

Happy learning (!)

Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!

n=1: &
Assume + ... + = for n=k
Let n=k+1:
+ ... + +
// Apply hypothesis to all but the last term
= +
// combine
=
// Here you can attempt to factor, or, you can see if (1/3)(k+1)(4(k+1)^2-1)
// expands to the same expression... I will do the latter approach...

=
=
=
=
They match, the hypothesis is also true for n=k+1, DONE.

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