Can someone help me derive the formula for the sum of the first n cubes from the formula for the sum of the first n integers that elucidates the reason why the former is the square of the latter?
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[tex](n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1[/tex] by the Binomial theorem.
Sum BOTH sides, and the LHS should get (N+1)^4 actually. Do you see how the RHS will be expressions in terms of the first N integers to the power of 3, 2, 1 and 0? You already know the cases for 0, 1 and 2, now you can work out 3.