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# Algebra

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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

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Successive  sums  produced

3        11       26       50       85

8        15      24        35            [   successive  differences  ]

7        9        11

2        2

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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   Apr 17, 2022