SOLUTION: Calculate the sum: 1/4(1^3) + 1/9((1^3) + (2^3)) + 1/16((1^3) + (2^3) + (3^3)) +...+ 1/2704((1^3) + (2^3) + (3^3) +...+ (51^3))

Algebra ->  Sequences-and-series -> SOLUTION: Calculate the sum: 1/4(1^3) + 1/9((1^3) + (2^3)) + 1/16((1^3) + (2^3) + (3^3)) +...+ 1/2704((1^3) + (2^3) + (3^3) +...+ (51^3))      Log On


   

Question 1191490: Calculate the sum:
1/4(1^3) + 1/9((1^3) + (2^3)) + 1/16((1^3) + (2^3) + (3^3)) +...+ 1/2704((1^3) + (2^3) + (3^3) +...+ (51^3))
Answer by greenestamps(13334) About Me  (Show Source):
You can put this solution on YOUR website!


The fractional part of the n-th term is

1%2F%28n%2B1%29%5E2

The other factor in the n-th term is the sum of the cubes of the first n integers:

sum%28%282n-1%29%5E3%2C1%2Cn%29=%28%28n%28n%2B1%29%29%2F2%29%5E2

So the nth term in the sequence is



There are 51 terms in the sequence, so the sum is

%28sum%28n%5E2%2C1%2C51%29%2F4%29

Use the formula for the sum of the first n squares to get the final answer.



ANSWER: 22763/4