SOLUTION: SUM THIS SERIES: 1^3+(1^3+2^3)+(1^3+2^3+3^3)+.......(1^3+2^3+...50^3)

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Question 1054887: SUM THIS SERIES:
1^3+(1^3+2^3)+(1^3+2^3+3^3)+.......(1^3+2^3+...50^3)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
SUM THIS SERIES:
1^3+(1^3+2^3)+(1^3+2^3+3^3)+.......(1^3+2^3+...50^3)


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:

%28n%28n%2B1%29%2F2%29%5E2

So we are looking for this:

sum%28%28%28n%28n%2B1%29%2F2%29%5E2%29%2Cn=1%2C50%29%22%22=%22%22

sum%28%28%28n%5E4%2B2n%5E3%2Bn%5E2%29%2F4%29%2Cn=1%2C50%29%22%22=%22%22

expr%281%2F4%29sum%28%28n%5E4%2B2n%5E3%2Bn%5E2%29%2Cn=1%2C50%29%22%22=%22%22

%22%22=%22%22

%22%22=%22%22

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 

%28n%282n%2B1%29%28n%2B1%29%283n%5E2%2B3n-1%29%29%2F30

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

sum of the first n squares
 %28n%28n%2B1%29%282n%2B1%29%29%2F6 

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  

%28n%28n%2B1%29%2F2%29%5E2
 
Substituting for the summations

%22%22=%22%22

which simplifies to

%28n%283n%5E4%2B15n%5E3%2B25n%5E2%2B15n%2B2%29%29%2F60 

which actually factors as

%28n%28n%2B1%29%28n%2B2%29%283n%5E2%2B6n%2B1%29%29%2F60

although that factorization isn't necessary

Substituting n = 50 gives

17240210

Edwin