# Square of the sum = Sum of the cubes

by rsala004
Tags: cubes, square
 P: 23 (1+2+....n)$$^{2}$$ = 1$$^{3}$$+2$$^{3}$$+...n$$^{3}$$ $$\frac{n^{2}(n+1)}{4}^{2}$$ = 1$$^{3}$$+2$$^{3}$$+...n$$^{3}$$ how do you simplify the right side to show that they are equal?
 P: 361 A much better idea is to use induction to prove the first equation. Petek
P: 436
 Quote by rsala004 (1+2+....n)$$^{2}$$ = 1$$^{3}$$+2$$^{3}$$+...n$$^{3}$$ $$\frac{n^{2}(n+1)}{4}^{2}$$ = 1$$^{3}$$+2$$^{3}$$+...n$$^{3}$$ how do you simplify the right side to show that they are equal?
What exactly is the question?
If the answer is to make you prove that square of sum equal to sum of cubes , you have to use mathematical induction to prove that sum is of the following formula $$\frac {n(n+1)}{2}$$ and once again use mathematical induction to prove that the sum of cubes is of the following forumula $$[\frac {n(n+1)}{2}]^{2}$$.Then only you can draw your conclusion.

 Mentor P: 15,202 Square of the sum = Sum of the cubes The question obviously is to prove (1+2+...+n)2 = 13+23+...n3. The obvious approach, as Petek noted, is to use induction.
Math
Emeritus
Thanks
PF Gold
P: 39,682
 Quote by rsala004 (1+2+....n)$$^{2}$$ = 1$$^{3}$$+2$$^{3}$$+...n$$^{3}$$ $$\frac{n^{2}(n+1)}{4}^{2}$$ = 1$$^{3}$$+2$$^{3}$$+...n$$^{3}$$ how do you simplify the right side to show that they are equal?
The standard simplification for the right side is
$$\frac{n^2(n+1)^2}{4}$$
And, as has been suggested several times now, can be proven by induction on n.
 P: 1 Hm, could someone please briefly explain induction? This thread has sparked my curiosity. I feel like I should know induction to n, but I don't :(