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Find a polynomial f(n) such that for all integers n>=1, we have

 

1*3 + 2*4 + 3*5 + ... + n*(n + 2) = f(n).

 

Write f(n) as a polynomial with terms in descending order or n.

 Apr 16, 2022
 #1
avatar+124594 
+1

Successive  sums  produced   

 

3        11       26       50       85

      8        15      24        35            [   successive  differences  ]

            7        9        11

                  2        2 

                        0

 

Sum of differences method

 

We have  3 rows of    non-zero   differences  

 

This will produce a 3rd power polynomial      an^2  + bn^2 + cn  + d

 

We have  the equations

 

a(1)^3  + b(1)^2  + c(1)  + d  = 3

a(2)^3  + b(2)^2  + c(2)   + d  =  11

a(3)^3 + b(3)^2  +  c(3)  +  d  = 26

a(4)^3 + b(4)^2  + c(4)   +  d  =  50

 

Simlifying we have

 

a  +  b +  c  +  d   = 3

8a + 4b + 2c + d =  11

27a  +9b + 3c + d  = 26

64a  + 16b + 4c + d =  50

 

This system is a little lengthy to solve (but not impossible !!! ) so I'll use a matrix  solver

 

a = 1/3

b = (3/2)

c = (7/6)

d  =  0

 

Our generating polynomial  is

 

(1/3)n^3  + (3/2)n^2  + (7/6)n

 

 

cool cool cool

 Apr 17, 2022

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