+845 
not quite sure about what the 3 infront of the symbol means please help
i have tried subbing in 3r but it doesnt work
+129852 The "3" is just a sum multiplier
We can use the sum of differences to find a polynomial
The frist 7 terms produced by the summation are :
6 6 6 12 30 66 126
0 0 6 18 36 60
0 6 12 18 24
6 6 6 6
We have 3 rows of non-zero differences.....so we have a third power polynomial of the form
an^3 + bn^2 + cn + d
And we have this system
a(1)^3 + b(1)^2 + c(1) + d = 6
a(2)^3 + b(2)^2 + c(2) + d = 6
a(3)^3 + b(3)^2 + c(3) + d = 6
a(4)^2 + b(4)^2 + c(4) + d = 12
a + b + c + d = 6
8a + 4b + 2c + d = 6
27a + 9b + 3c + d = 6
64a + 16b + 4c + d = 12
a = 1 b = -6 c = 11 d = 0
So.......we have the polynomial
n^3 - 6n^2 + 11n
We can show that this is equal to (n - 1)(n -2)(n -3) + 6
(n - 1)(n - 2) (n -3) + 6 =
(n^2 - 3n + 2) (n - 3) + 6 =
(n^3 - 3n^2 + 2n - 3n^2 + 9n - 6) + 6 =
(n^3 - 6n^2 + 11n - 6) + 6 =
n^3 - 6n^2 + 11n
PS, YEEEEEET......I can show you how to solve the system of equations if you want me to....
+845 thank you and
Please do, i think this is a different approach to the question than the one im taught but i really want to learn it
+129852 Yeah....I see that you might be able to use some type of induction to prove this, as well
Here's how to solve the system....it's lengthy but not that difficult
a + b + c + d = 6
8a + 4b + 2c + d = 6
27a + 9b + 3c + d = 6
64a + 16b + 4c + d = 12
Multiply the first equation by - 1 and add it to the second equation
7a + 3b + c = 0 (4)
Multiply the second equation by - 1 and add it to the third equation
19a + 5b + c = 0 (5)
Multiply the third equation by -1 and add it to the fourth equation
37a + 7b + c = 6 (6)
Multiply (4) by - 1 and add it to (5)
12a + 2b = 0 (7)
Multiply (5) by -1 and add it to (6)
18a + 2b = 6 (8)
Multiply (7) by -1 and add it to (8)
6a = 6
a = 1
Sub this into (7) to find b
12(1) + 2b = 0
2b = - 12
b = -6
Sub a, b into (4) to find c
7(1) + 3(-6) + c = 0
7 - 18 + c = 0
-11 + c = 0
c = 11
Sub a,b, c into the first equation to find d
1 - 6 + 11 + d = 6
-5 + 11 + d = 6
6 + d = 6
d = 0
So
a = 1 b = -6 c = 11 d = 0
+845 the method makes perfect sense and is fascinating but can u please explain where the 6 came from? just to clear up the little uncertainty thanks so much
+129852 OK
Note that the "n" in the polynomial represents the nth term in the summation
So.....the first term is "6"
So....we know that
a(n)^3 + b(n)^2 + c(n) + d = 6
So....in the first equation where we are trying to find the values for a, b, c and d we have
a(1)^3 + b(1)^2 + c(1) + d = 6
a + b + c + d = 6
But....we have already found a,b and c as 1, - 6 and 11
So
1 - 6 + 11 + d = 6
-5 + 11 + d = 6
6 + d = 6
So... d = 0
Does that make sense????
+6251 as an alternative approach we can use induction
\(P_1=3(-1)(-2) = (0)(-1)(-2)+6 = True\\ \text{show }P_n \Rightarrow P_{n+1}\\ 3\sum \limits_{r=1}^{n+1} ~(r-2)(r-3) = \\ \left(3\sum \limits_{r=1}^n(r-2)(r-3)\right)+3(n-1)(n-2) = \\ \left((n-1)(n-2)(n-3)+6\right) + 3(n-1)(n-2) = \\ (n-1)(n-2)((n-3)+3)+6 = \\ (n-1)(n-2)(n)+6 = \\ ((n+1)-1)((n+1)-2)((n+1)-3)+6 \Rightarrow P_{n+1}=True\)
