Perhaps you know that
![S_2 = \displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6](https://tex.z-dn.net/?f=S_2%20%3D%20%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5En%20k%5E2%20%3D%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D6)
and
![S_3 = \displaystyle\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}4](https://tex.z-dn.net/?f=S_3%20%3D%20%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5En%20k%5E3%20%3D%20%5Cfrac%7Bn%5E2%28n%2B1%29%5E2%7D4)
Then the problem is trivial, since
![\displaystyle\sum_{k=1}^n k^2(k+1) = S_2 + S_3 \\\\ = \frac{2n(n+1)(2n+1)+3n^2(n+1)^2}{12} \\\\ = \frac{n(n+1)\big((2(2n+1)+3n(n+1)\big)}{12} \\\\ = \frac{n(n+1)\big(4n+2+3n^2+3n\big)}{12} \\\\ = \frac{n(n+1)(3n^2+7n+2)}{12} \\\\ = \frac{n(n+1)(3n+1)(n+2)}{12}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%3D1%7D%5En%20k%5E2%28k%2B1%29%20%3D%20S_2%20%2B%20S_3%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B2n%28n%2B1%29%282n%2B1%29%2B3n%5E2%28n%2B1%29%5E2%7D%7B12%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7Bn%28n%2B1%29%5Cbig%28%282%282n%2B1%29%2B3n%28n%2B1%29%5Cbig%29%7D%7B12%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7Bn%28n%2B1%29%5Cbig%284n%2B2%2B3n%5E2%2B3n%5Cbig%29%7D%7B12%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7Bn%28n%2B1%29%283n%5E2%2B7n%2B2%29%7D%7B12%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7Bn%28n%2B1%29%283n%2B1%29%28n%2B2%29%7D%7B12%7D)
Then
![12\bigg(1^2\cdot2+2^2\cdot3+3^2\cdot4+\cdots+n^2(n+1)\bigg) = n(n+1)(n+2)(3n+1)](https://tex.z-dn.net/?f=12%5Cbigg%281%5E2%5Ccdot2%2B2%5E2%5Ccdot3%2B3%5E2%5Ccdot4%2B%5Ccdots%2Bn%5E2%28n%2B1%29%5Cbigg%29%20%3D%20n%28n%2B1%29%28n%2B2%29%283n%2B1%29)
so that <em>a</em> = 3 and <em>b</em> = 1.
Correct answer is
113.4 = x(18)
X = 6.3
Answer:
(D) Each year,the machinery loses a value of 4,500
When it is 3 years old,it is worth a little over 35K ,and when it is 6 years old,its worth about 22K.
This info means A,C and B is wrong b/c it loses too little value per year.
Answer:
i dont realy understand but i think it's 4