I have a mathematical induction question and I know what I need to do just not how to do it.
The question is:
Prove the equality of:
$$(1 + 2 + . . . + n)^2 = 1^3 + 2^3 . . . + n^3$$
Base case:
$$(1 + 2)^2 = 1^3 + 2^3\\
(3)^2 = 1 + 8\\
9 = 9$$
and I understand I have to get the sides to equal each other though I'm not sure how to do that:
I use this:
$$(1 + . . . + n + (n + 1))^2 = 1^3 + . . . + n^3 + (n + 1)^3$$
but i can't seem to factor anything in anyway to figure it out . . .
I've tried putting the $S(n)$ in the $S(n + 1)$:
$$(1 + . . . + n + (n + 1))^2 = (1 + . . . + n)^2 + (n + 1)^3$$
but its just getting the $-(n + 1)^3$ on the first side I can't figure out...
Any help would be amazing!!!!
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