Register to reply

Square of the sum = Sum of the cubes

by rsala004
Tags: cubes, square
Share this thread:
rsala004
#1
Nov6-09, 07:36 PM
P: 23
(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?
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
Petek
#2
Nov6-09, 09:28 PM
Petek's Avatar
P: 361
A much better idea is to use induction to prove the first equation.

Petek
icystrike
#3
Nov13-09, 06:15 AM
P: 436
Quote Quote by rsala004 View Post
(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?
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
#4
Nov13-09, 06:40 AM
Mentor
P: 15,167
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.
HallsofIvy
#5
Nov14-09, 04:17 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552
Quote Quote by rsala004 View Post
(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?
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
#6
Nov14-09, 01:08 PM
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 :(
CRGreathouse
#7
Nov14-09, 02:33 PM
Sci Advisor
HW Helper
P: 3,684
Quote Quote by prezjordan View Post
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
#8
Nov14-09, 02:42 PM
Mentor
P: 15,167
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.


Register to reply

Related Discussions
Cubes and the Imagination General Math 16
Why do ice cubes stick together? General Physics 4
Sum of the first n cubes Calculus & Beyond Homework 3
Two Cubes in 3d Fun, Photos & Games 12
Square Cubes Linear & Abstract Algebra 7