Alternative for calculating the nth of quadratic sequenceGiven the quadratic sequence f ( n ) = 1 , 7 , 19 , 37 , ⋯To calculate the f(n) for n ≥ 1. f ( n ) = a n 2 + b n + cWe start with the general quadratic function, then sub in for n := 1 , 2 and 3 f ( 1 ) = a + b + c f ( 2 ) = 4 a + 2 b + c f ( 3 ) = 9 a + 3 b + cNow solve the simultaneous equations (1) a + b + c = 1 (2) 4 a + 2 b + c = 7 (3) 9 a + 3 b + c = 19 ( 2 ) − ( 1 ) and ( 3 ) − ( 2 ) (4) 3 a + b = 6 (5) 5 a + b = 12 ( 5 ) − ( 4 ) a = 3 b = − 3 c = 1 f ( n ) = 3 n 2 − 3 n + 1This method is very long. Is there another easy of calculating the f(n)?

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Alternative for calculating the nth of quadratic sequenceGiven the quadratic sequence  f  (  n  )  =  1  ,  7  ,  19  ,  37  ,  ⋯To calculate the f(n) for   n  ≥  1.  f  (  n  )  =  a      n    2    +  b  n  +  cWe start with the general quadratic function, then sub in for   n  :=  1  ,  2 and 3  f  (  1  )  =  a  +  b  +  c  f  (  2  )  =  4  a  +  2  b  +  c  f  (  3  )  =  9  a  +  3  b  +  cNow solve the simultaneous equations                    (1)                    a        +        b        +        c        =        1                                (2)                    4        a        +        2        b        +        c        =        7                                (3)                    9        a        +        3        b        +        c        =        19              (  2  )  −  (  1  ) and   (  3  )  −  (  2  )                    (4)                    3        a        +        b        =        6                                (5)                    5        a        +        b        =        12              (  5  )  −  (  4  )  a  =  3  b  =  −  3  c  =  1  f  (  n  )  =  3      n    2    −  3  n  +  1This method is very long. Is there another easy of calculating the f(n)?

Phiplyrhypelw0 · Accepted Answer

Step 1You know   f  (  1  )  =  1    ⟹    f  (  1  )  −  1  =  0    ⟹    f  (  x  )  −  1 has a root at   x  =  1Now,  f  (  x  )  −  1  =  a  (  x  −  1  )  (  x  −  b  )Put   x  =  2,   a  (  2  −  b  )  =  6Put   x  =  3,   a  ∗  2  ∗  (  3  −  b  )  =  18Step 2Divide both equation, we get   b  =  0 and   a  =  3  f  (  x  )  −  1  =  3  ∗  (  x  −  1  )  x    ⟹    f  (  x  )  =  3      x    2    −  3  x  +  1

Demarion Ortega · Accepted Answer

Step 1Another standard way is to calculate a difference scheme and then to work backwards:                    0                          1                          2                          3                          4                                        1                          7                          19                          37                                              6                          12                          18                                                          6                          6                          ⇒                          0                          1                          2                          3                          4                                      c                =                              1                                  1                          7                          19                          37                                      a          +          b                =                                    0                                  6                          12                          18                                            2          a                =                                          6                                  6                          6                        Step 2  ⇒  a  =  3  ,    b  =  −  3  ,    c  =  1  ⇒  f  (  n  )  =  3      n    2    −  3  n  +  1

Alternative for calculating the nth of quadratic sequence. Given the quadratic sequence f(n)=1,7,19,37,...

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