Deriving the Formula for Sum of Cubes from Sum of Integers

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Sum
ehrenfest
Messages
2,001
Reaction score
1

Homework Statement


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?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Well you could consider the sum from n=1 to N of (n+1)^4-n^4 and then expand using the method of differences
 
You mean use the fact that it telescopes?

Then you just get N^4-1? How does that help?
 
(n+1)^4 - n^4 = 4n^3 + 6n^2 + 4n + 1 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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top