let-p-x-x-6-ax-5-bx-4-cx-3-dx-2-ex-f-be-a-polynomial-function-such-that-p-1-1-p-2-2-p-3-3-p-4-4-p-5-5-p-6-6-then-find-p-7-

Question Number 175624 by infinityaction last updated on 04/Sep/22

let p (x) = x^{6} + {ax}^{5} + {bx}^{4} + {cx}^{3} + {dx}^{2} + ex + f

be a polynomial function such

that p (1) = 1; p (2) = 2; p (3) = 3

p (4) = 4; p (5) = 5; p (6) = 6 then

find p (7) = ?

Answered by cortano1 last updated on 04/Sep/22

p (x) = (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) + x

p (7) = 6.5.4.3.2.1 + 7 = 6! + 7

= 727

Commented by BaliramKumar last updated on 05/Sep/22

p (n) = ? [n > 6]

Answered by leodera last updated on 04/Sep/22

p (x) = (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) + x

p (7) = 6! + 7 = 720 + 7 = 727

Answered by floor(10²Eta[1]) last updated on 04/Sep/22

p (1) = 1 \Rightarrow p (x) = (x - 1) a (x) + 1

p (2) = a (2) + 1 = 2 \Rightarrow a (2) = 1 \Rightarrow a (x) = (x - 2) b (x) + 1

p (x) = (x - 1) ((x - 2) b (x) + 1) + 1

p (3) = 2 (b (3) + 1) + 1 = 3 \Rightarrow b (3) = 0 \Rightarrow b (x) = (x - 3) c (x)

p (x) = (x - 1) ((x - 2) (x - 3) c (x) + 1) + 1

p (4) = 3 (2 c (4) + 1) + 1 = 4 \Rightarrow c (4) = 0 \Rightarrow c (x) = (x - 4) d (x)

\Rightarrow p (x) = (x - 1) [(x - 2) (x - 3) (x - 4) d (x) + 1) + 1

p (5) = 4 [3 . 2 d (5) + 1] + 1 = 5 \Rightarrow d (5) = 0 \Rightarrow d (x) = (x - 5) e (x)

p (x) = (x - 1) [(x - 2) (x - 3) (x - 4) (x - 5) e (x) + 1] + 1

p (6) = 5 [4.3 . 2 e (6) + 1] + 1 = 6 \Rightarrow e (6) = 0 \Rightarrow e (x) = (x - 6) C

\Rightarrow p (x) = (x - 1) [(x - 2) (x - 3) (x - 4) (x - 5) (x - 6) C + 1] + 1

= (x - 1) (x - 2) (x - 3) (x - 4) (x - 5) (x - 6) C + x

p (7) = 6! C + 7 = 720 C + 7, C \in R^{*}

but C = 1 because p (x) = x^{6} + {ax}^{5} + \dots

\Rightarrow p (7) = 727

Commented by infinityaction last updated on 04/Sep/22

t h a n k s s i r