Question 1054887
SUM THIS SERIES:
1^3+(1^3+2^3)+(1^3+2^3+3^3)+.......(1^3+2^3+...50^3)
<pre><b>

We can find all the formulas we need here:

http://godplaysdice.blogspot.com/2008/12/how-could-you-guess-formula-for-sum-of.html

The nth term is the sum of the first n cubes of the positive
integers:

We look up the formula for the sum of the first n cubes of 
positive integers:

{{{(n(n+1)/2)^2}}}

So we are looking for this:

{{{sum(((n(n+1)/2)^2),n=1,50)}}}{{{""=""}}}

{{{sum(((n^4+2n^3+n^2)/4),n=1,50)}}}{{{""=""}}}

{{{expr(1/4)sum((n^4+2n^3+n^2),n=1,50)}}}{{{""=""}}}

{{{expr(1/4)(sum((n^4),n=1,50)+
sum((2n^3),n=1,50)+
sum((n^2),n=1,50))}}}{{{""=""}}}

{{{expr(1/4)(sum((n^4),n=1,50)+
2sum((n^3),n=1,50)+
sum((n^2),n=1,50))}}}{{{""=""}}}

Now we must look up the formula for the sum
of the first n 4th powers and squares of positive
integers:

The sum of the first n 4th powers of positive integers is 

{{{(n(2n+1)(n+1)(3n^2+3n-1))/30}}}

The sum of the first n squares of powers of positive integers is 

sum of the first n squares
 {{{(n(n+1)(2n+1))/6}}} 

We have already looked up the formula for the sum of the first 
n cubes of powers of positive integers 
sum of the first n cubes, which again is  

{{{(n(n+1)/2)^2}}}
 
Substituting for the summations

{{{expr(1/4)( ( n(2n+1)(n+1)(3n^2+3n-1) )/30+
2( (n(n+1))/2 )^2+
(n(n+1)(2n+1))/6 )}}}{{{""=""}}}

which simplifies to

{{{(n(3n^4+15n^3+25n^2+15n+2))/60}}} 

which actually factors as

{{{(n(n+1)(n+2)(3n^2+6n+1))/60}}}

although that factorization isn't necessary

Substituting n = 50 gives

17240210

Edwin</pre>