# Square of the sum = Sum of the cubes

## Main Question or Discussion Point

(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?

## Answers and Replies

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A much better idea is to use induction to prove the first equation.

Petek

(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.

D H
Staff Emeritus
The question obviously is to prove (1+2+...+n)2 = 13+23+...n3.

The obvious approach, as Petek noted, is to use induction.

HallsofIvy
Homework Helper
(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.

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 :(

CRGreathouse
Homework Helper
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 :(
Induction: if you can prove
1. f(1) is true
2. If f(n) is true, then f(n+1) is true
for any proposition f, then f(n) is true for all n.

D H
Staff Emeritus