Why does the cube of natural numbers equal to the square of the same numbers when they are added.?

Question

eg.(1to the third power+2 to the third power+3 to the third power)=(1+2+3) all squared

Answer 1

kay,here is the proof

1^3=3
1^3+2^3=9
1^3+2^3+3^3=36 etc

1,9,36,100,225,441 etc

using difference theory;

for n =1,2,3........
nth term=1/4n^4+1/2n^3+1/4n^2
=n^2(1/4n^2+1/2n+1/4)
=n^2(1/2n+1/2)^2
={ n(1/2n+1/2)}^2
=k, say

now,for n=1,
k=1
for n=2
k=3^2 =(1+2)^2
for n=3
k=6^2 =(1+2+3)^2
for n=4,
k=(10)^2=(1+2+3+4)^2
for n=n,
k=(1+2+3+......n)^2

but,it has been shown that k is the
nth term for the sequence

1,9,36,100,225,441 etc

therefore,

(1^3+2^3+3^3+.......n^3)
=(1+2+3+......n)^2

if you don't follow the reasoning,
please contact me

note:the sum of 1 to n is n/2(n+1)
k is simply the square of that

k is the sum of all the cubes
from 1 to n

Answer 2

All I'm doing is presenting essentially the same proof as a couple of previous answers, but I'll write it in a slightly different way.

Let C[n] be the the sum of the first n cubes of natural numbers.
Let Z[n] be the sum of the first first n natural numbers = n(n+1)/2
Let S[n] be Z[n] squared.

By inspection it is obvious that 1 = C1 = S1 so the equation is true for n=1


By definition C[n] = C[n-1] + n^3 (for n >1)

Also by definition Z[n] = Z[n-1] + n (for n>1)

So S[n] = (Z[n-1] + n)^2

= n^2 +2nZ[n-1] + Z[n-1]^2

= n^2 +2nZ[n-1] + S[n-1]

So

S[n-1] = S[n] - ( n^2 +2nZ[n-1] )

Using induction, we need to prove that if C[n-1] = S[n-1] then C[n] = S[n]

C[n] = S[n-1] +n^3 (by inductive hypothesis)

= S[n] - ( n^2 +2nZ[n-1] ) + n^3 .............. (1)

We also know that Z[n-1] = n(n-1)/2 so expression (1) equals

= S[n] - ( n^2 +2n2(n-1)/2) + n^3

= S[n] - ( n^2 +2n2(n-1)/2) + n^3

= S[n] - ( n^2 +n^3-n^2)) + n^3

= S[n]

So we have C[n] =S[n] for all values of n Q.E.D.

(More simply, 'why' ... 'because its a property of the way the natural numbers are defined').

Answer 3

There is a "proof" by picture in the source.

You can prove, using mathematical induction that

     1 + 2 + 3 + ... + n = n (n + 1) / 2

and

     1³ + 2³ + 3³ + ... + n³ = ¼ n² (n + 1)²

Answer 4

madrid, it's the natural numbers so you must start with 1, then go consecutively as far as you want and it does work.

Sorry, I don't have the proof right now but I'll work on it.

Answer 5

Dur...simple math!
Its bound to...its the CUBE of natural square!
Think about it!

Answer 6

i don't know