Square of the sum = Sum of the cubes


by rsala004
Tags: cubes, square
rsala004
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#1
Nov6-09, 07:36 PM
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(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?
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Petek
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#2
Nov6-09, 09:28 PM
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A much better idea is to use induction to prove the first equation.

Petek
icystrike
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#3
Nov13-09, 06:15 AM
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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
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#4
Nov13-09, 06:40 AM
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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
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#5
Nov14-09, 04:17 AM
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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
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#6
Nov14-09, 01:08 PM
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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
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#7
Nov14-09, 02:33 PM
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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
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#8
Nov14-09, 02:42 PM
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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.


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