+0  
 
0
40
10
avatar+622 

not quite sure about what the 3 infront of the symbol means please help 

i have tried subbing in 3r but it doesnt work 

 Dec 30, 2018
 #1
avatar+94209 
+2

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....

cool cool cool

 Dec 30, 2018
edited by CPhill  Dec 30, 2018
edited by CPhill  Dec 30, 2018
 #2
avatar+622 
+1

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 

YEEEEEET  Dec 30, 2018
 #3
avatar+94209 
+2

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

 

 

cool cool cool

CPhill  Dec 30, 2018
 #5
avatar+622 
0

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

YEEEEEET  Dec 30, 2018
 #7
avatar+94209 
+1

Which "6"  are you referring to   ????

 

 

cool cool cool

CPhill  Dec 30, 2018
 #8
avatar+622 
+1

the one in the equation to solve the system 

a+b+c+d = 6

YEEEEEET  Dec 30, 2018
 #9
avatar+94209 
+2

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????

 

 

cool cool cool

CPhill  Dec 30, 2018
 #10
avatar+622 
+1

yes perfect, thank you again for teaching me a new method 

YEEEEEET  Dec 30, 2018
 #4
avatar+3535 
+1

as an alternative approach we can use induction

 

\(P_1=3(-1)(-2) = (0)(-1)(-2)+6 = True\\ \text{show }P_n \Rightarrow P_{n+1}\\ 3\sum \limits_{r=1}^{n+1} ~(r-2)(r-3) = \\ \left(3\sum \limits_{r=1}^n(r-2)(r-3)\right)+3(n-1)(n-2) = \\ \left((n-1)(n-2)(n-3)+6\right) + 3(n-1)(n-2) = \\ (n-1)(n-2)((n-3)+3)+6 = \\ (n-1)(n-2)(n)+6 = \\ ((n+1)-1)((n+1)-2)((n+1)-3)+6 \Rightarrow P_{n+1}=True\)

.
 Dec 30, 2018
 #6
avatar+622 
+1

ah i see thank you!

YEEEEEET  Dec 30, 2018

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