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The sum
6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right)
is equal to a polynomial f(n) for all n \ge 1.

 

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

 Jun 18, 2024
 #1
avatar+129732 
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\(6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) \)

 

For the moment, disregard the "6"

 

We have

 

1 +  4 +  9 +  16  +  25 +  36

 1      5     14      30     55    91

    4      9     16      25     35

        5     7       9        11

            2      2        2

 

We have  three rows of non-zero differences

 

The generating polynomial willl be

an^3 + bn^2 + cn + d

 

We have this system

a + b + c + d  = 1

8a + 4b + 2c + d = 5

27a + 9b + 3c + d = 14

64a + 16b + 4 + d = 30

 

Solving this  produces

a = 1/3     b =  1/2  c  =1/6   d  = 0

 

The polynomial is

 

(1/3)n^3 + (1/2)n^2 + (1/6)n

 

Multiplying by 6  we get

 

f(n) = 2n^2 + 3n  + n

 

 

cool cool cool

 Jun 18, 2024

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