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 correct answer according to me is 3=12
<span>a) It is prime - 2, 3
b) It is odd number and less than 5 - 1, 3,
c) It is an even number - 2,4
d) It is 3
If a) and c) is true, then Number is 2. 2 is prime and 2 is even.
b) and c) is false.</span>
Answer:
90$
Step-by-step explanation:
First divide 6 by 2 to get 3 then multiply 30 by that value