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Find 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.

 Nov 27, 2020
 #1
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We have these summations.....using the  sum of differences

 

3           11           26          50           85

        8          15           24          35

             7             9            11

                   2              2

                          0

 

We have  3 rows  of non-zero  differences  until we get to a zero difference row

 

This will  give us a 3rd degree polynomial   ax^3  + bx^2 + cx  + d

 

We  have the following  system

a   +  b   +  c  +  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

 

Simplifying this, we get

 

a + b + c + d  = 3

8a +  4b + 2c + d =  11

27a + 9b + 3c + d =  26

64a  + 16b + 4c + d  = 50

 

This system isn't difficult to solve, just tedious......I used WolframAlpha  to get that

a= 1/3   b = 3/2  c  = 7/6   d  = 0

 

So our polynomial is

 

f(n)  = (1/3)n^3  +  (3/2)n^2  + (7/6)

 

 

cool cool cool                

 Nov 27, 2020

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