What is the sum of cubes in a series up to 'n' terms?

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SUMMARY

The discussion focuses on calculating the sum of cubes in a series up to 'n' terms, specifically the expression (1^3 / 1) + (1^3 + 2^3 / 1+2) + (1^3 + 2^3 + 3^3 / 1+2+3) + ... The relevant formulas provided include the summation of squares, sigma n^2 = n(n+1)(2n+1) / 6, and the summation of integers, sigma n = n(n+1) / 2. The correct approach involves recognizing that the numerator must sum cubes, leading to the realization that the series converges to triangular numbers, represented as 1, 3, 6, 10, 15, ... for the first few terms.

PREREQUISITES
  • Understanding of summation notation and series
  • Familiarity with the formulas for sigma notation, specifically sigma n^2 and sigma n
  • Basic knowledge of cubic functions and their properties
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the derivation of the formula for the sum of cubes: (n(n+1)/2)^2
  • Learn about triangular numbers and their properties
  • Explore the relationship between sums of powers and polynomial identities
  • Investigate the application of series in calculus and mathematical analysis
USEFUL FOR

Students studying mathematics, particularly those focusing on series and sequences, educators teaching algebraic concepts, and anyone interested in advanced summation techniques.

draotic
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Homework Statement


find the sum of the series to 'n' terms
(1^3 / 1 ) + (1^3 + 2^3 / 1+2 ) + (1^3 + 2^3 + 3^3 / 1+2+3 ) +...


Homework Equations



sigma n^2 = n(n+1)(2n+1) / 6
sigma n = n(n+1) / 2


The Attempt at a Solution


numerator = 1^3 + 2^3 + ... n^3
denominator = 1+2+3+...n
...
my answer comes out to be n(n+1) / 2 , but i think that's wrong...
please lead me
 
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That is the last term you are calculating. And anyway, the question says Sum of cubes (not squares) in numerator. If you evaluate first few terms, you get 1+3+6+10+15+...
I think you can work this out.
 

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