question dealing with squares of positive integers?

Question

suppose that you want to add the squares of positive integers starting with 1 and going up to n. this can b accomplished by the following formula: s(n)= 1/3n^3+1/2n^2+1/6n

a) make a table of both positive and negative integers. be sure to include all of the zeros of the function. (plz show all work)

b) for each of the following values, determine if they are in range of s(n). if so, identify the corresponding domain value. (plz show all work)
i. 92
ii. 385
iii. 179
iv. 204

thx a million!

Answer 1

a) First of all, the zeros of the function can be found by solving the equation for zero as follows:

1/3n^3+1/2n^2+1/6n = 0
factoring the 3rd order polynomial we get:
1/6n(n+1)(2n+1) = 0
therefore, the three zeros are:
1/6n=0 => n=0
n+1=0 => n=-1
2n+1=0 => n=-1/2 (Notice that -1/2 is not an integer, thus only two zeros, n=0 and -1, are answers.)

Then our table can be made just by plugging in various integer values of n into s(n) and it looks like:

n: -4, -3, -2, -1, 0, 1, 2, 3, 4
s(n): -14, -5, -1, 0, 0, 1, 5, 14, 30

b) First of all, for a number to be in the RANGE of s(n), there must exist a positive integer value n such that when plugged into the equation s(n), will give you that number. Now there are two ways to go about this part.

One is the bull headed way which requires you to solve each equation for n when s(n) is equal to one of the values given below:

Ex.

i. 92 = 1/3n^3+1/2n^2+1/6n
0 = 1/3n^3+1/2n^2+1/6n -92

At this point, unless you feel like solving 3rd degree polynomials, (not undoable but very tedious) its actually faster to do a guess and check. The key is that n must be positive integers so we need only start plugging in values of 1,2,3, etc....

i. when n = 6, s(n) =91. when n=7, s(n) =140. Therefore, s(n) cannot equal 92 (since there's no other integers between 6 and 7) and 92 is NOT in the range of s(n).

ii. when n = 10, s(n) =385. Therefore 385 is in the range of s(n) and the corresponding domain value is n = 10.

iii. when n=7, s(n)=140. when n=8, s(n) =204. Using the same logic as in part i, we can say that s(n) cannot equal 179 therefore it is not in the range of s(n).

iv. we just solved in part iii that when n=8, s(n)=204. Therefore, 204 is in the range of s(n) and the corresponding domain value is n=8.

If your teacher really wants you to solve a 3rd degree polynomial to find an answer (I doubt it), here's a good reference:
http://www.themathpage.com/aPreCalc/factor-theorem.htm
In the interest of time, and because I'm rather lazy, I'd rather not do it that way.

Answer 2

a) make a table of both positive and negative integers. be sure to include all of the zeros of the function. (plz show all work)

s(n)= 1/3 n³ +1/2 n²+1/6 n
= n/6(2n² + 3n + 1)
=n/6 (n + 1)(2n + 1)

So zeros at n = 0, -½, -1

n ...... s(n)
-8 ... -140
-7 ..... -91
-6 ..... -55
-5 ..... -30
-4 ..... -14
-3 ....... -5
-2 ....... -1
-1 ........ 0
-0.5 ..... 0
0 ......... 0
1 ......... 1
2 ......... 5
3 ....... 14
4 ....... 30
5 ....... 55
6 ....... 91
7 ..... 140
8 ..... 204
9 ..... 285
10 ... 385
11 ... 506

b) for each of the following values, determine if they are in range of s(n). if so, identify the corresponding domain value. (plz show all work)
i. 92 ← No
ii. 385 ← Yes when n = 10
iii. 179 ← No
iv. 204 ← Yes when n = 8

Answer 3

a) Just do it.
b) iv. s(8)=204 and ii. s(10)=385 are in the domain

Answer 4

zeros = { 0 , -1 , -1/2}
at this link
http://img226.imageshack.us/img226/9150/untitledwf8.jpg

Answer 5

Could you rephrase that please?