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Square of the sum = Sum of the cubes 
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#1
Nov609, 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? 


#2
Nov609, 09:28 PM

P: 361

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


#3
Nov1309, 06:15 AM

P: 436

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. 


#4
Nov1309, 06:40 AM

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P: 15,202

Square of the sum = Sum of the cubes
The question obviously is to prove (1+2+...+n)^{2} = 1^{3}+2^{3}+...n^{3}.
The obvious approach, as Petek noted, is to use induction. 


#5
Nov1409, 04:17 AM

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Thanks
PF Gold
P: 39,682

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


#6
Nov1409, 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 :(



#7
Nov1409, 02:33 PM

Sci Advisor
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P: 3,682

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
Nov1409, 02:42 PM

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P: 15,202

The basic idea behind using induction to prove some indexed relationship f_{n} is
The problem at hand is to show that 1+2+...+n)^{2} = 1^{3}+2^{3}+...n^{3}. This equality is trivially true for the case n=1; it reduces to 1^{2}=1^{3}=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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