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:
![\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Di%5E3%20%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%5E2)
And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2
![\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%5E2)
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:

The 5 in 59 is tens and the 5 in .085 is thousandths.
Good luck!
Answer:
5
Step-by-step explanation:
8 = y + 3
5 + 3 = 8
Therefore y = 5
Answer:
C
Step-by-step explanation:
Hope it helps, Good Luck!!!!