+0

# summation series

0
173
10
+846

i have tried subbing in 3r but it doesnt work

Dec 30, 2018

#1
+101804
+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....

Dec 30, 2018
edited by CPhill  Dec 30, 2018
edited by CPhill  Dec 30, 2018
#2
+846
+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
+101804
+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

CPhill  Dec 30, 2018
#5
+846
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
+101804
+1

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

CPhill  Dec 30, 2018
#8
+846
+1

the one in the equation to solve the system

a+b+c+d = 6

YEEEEEET  Dec 30, 2018
#9
+101804
+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????

CPhill  Dec 30, 2018
#10
+846
+1

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

YEEEEEET  Dec 30, 2018
#4
+5225
+2

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
+846
+1

ah i see thank you!

YEEEEEET  Dec 30, 2018