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 in part (a). 13 23 33 n3

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:

[tex] \lim_{n\to\infty} \sum_{n=1}^{\infty} i^3[/tex]

We can express this formula like this:

[tex] \lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2[/tex]

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

[tex] \lim_{n\to\infty} [\frac{n(n+1)}{2}]^2[/tex]

If we operate and we take out the 1/4 as a factor we got this:

[tex] \lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}[/tex]

We can cancel [tex]n^2[/tex] and we got

[tex] \lim_{n\to\infty} \frac{(n+1)^2}{n^2}[/tex]

We can reorder the terms like this:

[tex] \lim_{n\to\infty} (\frac{n+1}{n})^2[/tex]

We can do some algebra and we got:

[tex] \lim_{n\to\infty} (1+\frac{1}{n})^2 [/tex]

We can solve the square and we got:

[tex] \lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})[/tex]

And when we apply the limit we got that:

[tex]\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1 [/tex]