+0

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

+3
351
6

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

Guest Aug 10, 2014

#4
+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
Sort:

#1
+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
+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
+92251
+3

Why is that true Alan?

Is there something obvious that I am missing?

Melody  Aug 10, 2014
#4
+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
#5
+92251
+3

Thanks Alan,

I'll spend more time in it.

Melody  Aug 10, 2014
#6
+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

### 16 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details