CPhill

https://web2.0calc.com/questions/help-geometry_57877

https://web2.0calc.com/questions/geometry_47534

6(12)^4 + 0 (12)^3 + 3(12)^2 + 2(12) + 9   =  124881₁₀

9(16)^2  + 8(16) + 7  = 2439₁₀

35 / 8 =  4 R 3

4 / 8  =   0 R 4

35₁₀  = 43₈

\(6\left( 1\cdot1 + 2\cdot2 + \dots + n(n) \right) \)

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

https://web2.0calc.com/questions/floor-function_46

xy + x = (3x + 2y + z) / 2

2 (xy + x)  = 3x + 2y + z

2xy + 2x =  3x + 2y + z

2xy - x = 2y + z

x ( 2y -1)  = 2y + z

x = ( 2y + z)  / ( 2y  -1)

This is  impossible...a cubic polynomial has - at most - three real roots.....this polynomial has four.....at a minimum, it must be a quartic

2

2 [   2    - 9    13    k  ]

               4  - 10    6

    __________________

      2     - 5      3   k + 6

k + 6 = 0

k = -6

The  remaining polynomial  is

2x^2  - 5x  + 3  =  0

(2x - 3) ( x - 1)  = 0

The other two roots   are  3/2  and 1