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1^3 + 2^3 + 3^3 + 4^3 + 5^3 +.... 2014^3

Guest Aug 10, 2014

Best Answer 

 #4
avatar+26642 
+8

It's not at all obvious!  But have a look at https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes for a proof. 

Alan  Aug 10, 2014
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6+0 Answers

 #1
avatar+92251 
+8

You can probably do this on the web2 calc as well but i don't know how.

anyway, I got you the answer. 

 

Melody  Aug 10, 2014
 #2
avatar+26642 
+8

Can find this from:

$$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$

$${\frac{{\left({\mathtt{2\,014}}{\mathtt{\,\times\,}}{\mathtt{2\,015}}\right)}^{{\mathtt{2}}}}{{\mathtt{4}}}} = {\mathtt{4\,117\,267\,101\,025}}$$

Alan  Aug 10, 2014
 #3
avatar+92251 
+3

Why is that true Alan?

Is there something obvious that I am missing?

Melody  Aug 10, 2014
 #4
avatar+26642 
+8
Best Answer

It's not at all obvious!  But have a look at https://proofwiki.org/wiki/Sum_of_Sequence_of_Cubes for a proof. 

Alan  Aug 10, 2014
 #5
avatar+92251 
+3

Thanks Alan,

I'll spend more time in it. 

Melody  Aug 10, 2014
 #6
avatar+19207 
0

1^3 + 2^3 + 3^3 + 4^3 + 5^3 +.... 2014^3  

 

$$s_1 = 1+2+3+ ... + n$$

$$\begin{array}{llcl}
& (1+i)^2 &=& i^2+2i+1 \\
& (1+i)^2 - 1 &=& 1+2i \\
\hline
i=1 & \not{2^2}-1^2 &=& 1+ 2*1\\
i=2 & \not{3^2}-\not{2^2} &=& 1+ 2*2 \\
i=3 & \not{4^2}-\not{3^2} &=& 1+ 2*3 \\
... &... &...& ... \\
i=n & (1+n)^2-\not{n^2} &=& 1+2n\\
\hline
\Sigma & (1+n)^2-1&=&n+2*(\underbrace{1+2+3+...+n}_{s_1})
\end{array}\\
(1+n)^2-1&=&n+2s_1\\
n+n^2=2s_1\\
\boxed{s_1=\dfrac{n(n+1)}{2}}$$

 

$$s_2= 1^2+2^2+3^2+ ... + n^2$$

$$\begin{array}{llcl}
& (1+i)^3 &=& i^3+3i^2+3i+1 \\
& (1+i)^3 - i^3 &=& 1+3i^2+3i \\
\hline
i=1 & \not{2^3}-1^3 &=& 1+ 3*1^2+ 3*1 \\
i=2 & \not{3^3}-\not{2^3} &=& 1+ 3*2^2+ 3*2 \\
i=3 & \not{4^3}-\not{3^3} &=& 1+ 3*3^2+ 3*3 \\
... &... &...& ... \\
i=n & (1+n)^3-\not{n^3} &=& 1+ 3*n^2+ 3n \\
\hline
\Sigma & (1+n)^3-1&=&n+3*(\underbrace{1^2+2^2+3^2+...+n^2}_{s_2})+3*(\underbrace{1+2+3+...+n}_{s_1})
\end{array}\\
(1+n)^3-1 = n+3s_2+3s_1\\
(1+n)^3-1 = n+3s_2+3*\frac{n*(n+1)}{2}\\
s_2=n*\left( \frac{1}{6}+\frac{n}{2}+\frac{n^2}{3}\right)\\
s_2=\frac{n}{6}*(1+3n+2n^2)\\
\boxed{s_2=\dfrac{n}{6}(n+1)(2n+1) }$$

 

$$s_3= 1^3+2^3+3^3+ ... + n^3$$

$$\begin{array}{llcl}
& (1+i)^4 &=& i^4+4i^3+6i^2+4i+1 \\
& (1+i)^4 - i^4 &=& 1+4i^3+6i^2+4i\\
\hline
i=1 & \not{2^4}-1^4 &=& 1+ 4*1^3+ 6*1^2+ 4*1 \\
i=2 & \not{3^4}-\not{2^4} &=& 1+ 4*2^3+ 6*2^2+ 4*2 \\
i=3 & \not{4^4}-\not{3^4} &=& 1+ 4*3^3+ 6*3^2+ 4*3 \\
... &... &...& ... \\
i=n & (1+n)^4-\not{n^4} &=& 1+ 4*n^3+ 6*n^2+ 4*n\\
\hline
\Sigma & (1+n)^4-1&=&n+4*(\underbrace{1^3+2^3+3^3+...+n^3}_{s_3})
+6*(\underbrace{1^2+2^2+3^2+...+n^2}_{s_2}) +4*(\underbrace{1+2+3+...+n}_{s_1})
\end{array}\\
(1+n)^4-1 = n+4s_3+6s_2+4s_1\\
(1+n)^4-1 = n+4s_3+6*\frac{n}{6}(n+1)(2n+1)+4*\frac{n*(n+1)}{2}\\
4*s_3=n^4+4n^3+6n^2+4n-n-2n^3-3n^2-n-2n^2-2n\\
4*s_3=n^4+2n^3+n^2\\
\boxed{s_3=\dfrac{n^2*(n+1)^2}{4}}$$

$$1^3 + 2^3 + 3^3 + 4^3 + 5^3 +.... 2014^3 = \left(
\dfrac{2014*2015}{2}
\right)^2 \\
= 2029105^2\\
=4\;117\;267\;101\;025$$

heureka  Aug 21, 2014

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