Square of the sum = Sum of the cubes

rsala004

(1+2+....n)[tex]^{2}[/tex] = 1[tex]^{3}[/tex]+2[tex]^{3}[/tex]+...n[tex]^{3}[/tex]

[tex]\frac{n^{2}(n+1)}{4}^{2}[/tex] = 1[tex]^{3}[/tex]+2[tex]^{3}[/tex]+...n[tex]^{3}[/tex]

how do you simplify the right side to show that they are equal?

Petek

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

Petek

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 [tex]\frac {n(n+1)}{2}[/tex] and once again use mathematical induction to prove that the sum of cubes is of the following forumula [tex][\frac {n(n+1)}{2}]^{2}[/tex].Then only you can draw your conclusion.

D H

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

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

HallsofIvy

The standard simplification for the right side is
[tex]\frac{n^2(n+1)^2}{4}[/tex]
And, as has been suggested several times now, can be proven by induction on n.

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 :(

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.

D H

The basic idea behind using induction to prove some indexed relationship f_n 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.