Respuesta :

Answer:

  third differences are all 6a

Step-by-step explanation:

Substituting for x, we find the sequence values to be ...

  (a+b+c+d), (8a+4b+2c+d), (27a+9b+3c+d), (64a+16b+4c+d), (125a+25b+5c+d),

(216a+36b+6c+d)

Then first differences are ...

  (7a+3b+c), (19a+5b+c), (37a+7b+c), (61a+9b+c), (91a+11b+c)

Second differences are ...

  (12a+2b), (18a+2b), (24a+2b), (30a+2b)

And the third differences are ...

  (6a), (6a), (6a) . . . . . constant

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Additional comment

These results can help you write a polynomial relation up to 3rd degree for a sequence that has constant differences at any level up to 3rd differences.

Of course, we find the differences by subtracting each term from the one following.

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