# Square of the sum = Sum of the cubes

1. Nov 6, 2009

### 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?

2. Nov 6, 2009

### Petek

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

Petek

3. Nov 13, 2009

### icystrike

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.

4. Nov 13, 2009

### 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.

5. Nov 14, 2009

### HallsofIvy

Staff Emeritus
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.

6. Nov 14, 2009

### prezjordan

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 :(

7. Nov 14, 2009

### CRGreathouse

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.

8. Nov 14, 2009

### D H

Staff Emeritus
The basic idea behind using induction to prove some indexed relationship fn is
• Show that the relationship is true for some particular value of n0.
• Show that if the relationship is true for n0, n0+1,n0+2, ..., N, then it is true for N+1.
By induction, these two items mean that the relationship is true for all integers greater than or equal to n0.

The problem at hand is to show that 1+2+...+n)2 = 13+23+...n3. This equality is trivially true for the case n=1; it reduces to 12=13=1. The trick then is to show that if the relationship is true for some particular n, then it is also true for n+1.