A pro-athlete is offered an eight-year
1 answer:
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Answer:
And when we apply the limit we got that:
Step-by-step explanation:
Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"
We have the following formula in order to find the sum of cubes:
We can express this formula like this:
And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2
If we operate and we take out the 1/4 as a factor we got this:
We can cancel and we got
We can reorder the terms like this:
We can do some algebra and we got:
We can solve the square and we got:
And when we apply the limit we got that: