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for the sequence 0, 6, 16,30, 48,..., what is the 40th term

 Sep 14, 2014

Best Answer 

 #1
avatar+129840 
+5

 

We can find a generating polynomial by the sum-of-differences technique

0, 6, 16, 30, 48

 6  10   14  18

   4    4     4

      0     0

There are three differences required to get to the "0" row

This means that we have a possible 3rd degree polynomial with at least some of the coefficients ≠ 0

P(x) = ax3 + bx2 + cx + d

 When x = 1, we have   a + b + c + d = 0             (1)

 When x = 2, we have  8a + 4b + 2c + d = 6         (2)             

 When x = 3, we have  27a + 9b + 3c + d = 16      (3)

 When x = 4, we have  64a + 16b + 4c + d = 30     (4)

 

Subtracting (1) from (2) we have       7a + 3b + c = 6      (5)

Subtracting (2) from (3) we have      19a + 5b + c = 10    (6)

Subtracting (3) from (4) we have      37a + 7b + c = 14    (7)

 

Subtracting (5) from (6) we have     12a + 2b = 4      (8)

Subtracting (6) from (7) we have     18a + 2b = 4      (9)      

And subtracting  (8) from (9)  we get 

6a = 0   →   a =  0        

And back-substituting  to find the other coefficients, we have 

b= 2   →   c = 0  →  d = -2

And our polynomial turns out to be the quadratic

P(x) = 2x2 - 2

And the 40th term is given by

P(40 ) = 2(40)2 - 2   =   3198

 

 

 Sep 14, 2014
 #1
avatar+129840 
+5
Best Answer

 

We can find a generating polynomial by the sum-of-differences technique

0, 6, 16, 30, 48

 6  10   14  18

   4    4     4

      0     0

There are three differences required to get to the "0" row

This means that we have a possible 3rd degree polynomial with at least some of the coefficients ≠ 0

P(x) = ax3 + bx2 + cx + d

 When x = 1, we have   a + b + c + d = 0             (1)

 When x = 2, we have  8a + 4b + 2c + d = 6         (2)             

 When x = 3, we have  27a + 9b + 3c + d = 16      (3)

 When x = 4, we have  64a + 16b + 4c + d = 30     (4)

 

Subtracting (1) from (2) we have       7a + 3b + c = 6      (5)

Subtracting (2) from (3) we have      19a + 5b + c = 10    (6)

Subtracting (3) from (4) we have      37a + 7b + c = 14    (7)

 

Subtracting (5) from (6) we have     12a + 2b = 4      (8)

Subtracting (6) from (7) we have     18a + 2b = 4      (9)      

And subtracting  (8) from (9)  we get 

6a = 0   →   a =  0        

And back-substituting  to find the other coefficients, we have 

b= 2   →   c = 0  →  d = -2

And our polynomial turns out to be the quadratic

P(x) = 2x2 - 2

And the 40th term is given by

P(40 ) = 2(40)2 - 2   =   3198

 

 

CPhill Sep 14, 2014

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