SOLUTION: Prove that 1^2 + 3^2 + . . . + (2n − 1)^2 =1/3(4n^3 − n) for all natural numbers n.

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Question 1023504: Prove that
1^2 + 3^2 + . . . + (2n − 1)^2 =1/3(4n^3 − n)
for all natural numbers n.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
1%5E2+%2B+3%5E2++ . . . + %282n+-+1%29%5E2+=%281%2F3%29%284n%5E3+-+n%29
Proof by induction.
For n = 1, the statement is trivially true.
Assume the statement is true for some natural number n = k,
or
1%5E2+%2B+3%5E2++ . . . + %282k+-+1%29%5E2+=%281%2F3%29%284k%5E3+-+k%29
Then by adding %282k%2B1%29%5E2 on both sides, we get
1%5E2+%2B+3%5E2++ . . . + %282k+-+1%29%5E2+%2B+%282k%2B1%29%5E2=%281%2F3%29%284k%5E3+-+k%29%2B+%282k%2B1%29%5E2
=, and the statement is proved...