Let-P-x-be-a-polynomial-function-of-degree-n-such-that-P-k-k-k-1-for-k-0-1-2-n-Then-P-n-1-is-equal-to-A-1-if-n-is-even-B-1-if-n-is-odd-C-n-n-2-if-n-is-even-

Question Number 117649 by Ar Brandon last updated on 13/Oct/20

Let P (x) be a polynomial function of degree n such that

P (k) = \frac{k}{k + 1} for k = 0, 1, 2, \dots, n . Then P (n + 1) is equal to

(A) - 1 if n is even (B) 1 if n is odd

(C) \frac{n}{n + 2} if n is even (D) \frac{n}{n + 2} if n is odd

Which among the four proposals is / are correct ?

Answered by prakash jain last updated on 13/Oct/20

Q (k) = (k + 1) P (k) - k \dots . . (A)

Q (0) = 0

Q (1), \dots . Q (n) = 0

Q (k) has zeros at 0, 1, 2, \dots, n

Q (k) = c k (k - 1) (k - 2) \dots . (k - n)

F r o m (A) Q (- 1) = 1

1 = c (- 1) (- 2) \dots . (- 1 - n)

c = \frac{{(- 1)}^{n + 1}}{(n + 1)!}

Q (n + 1) = \frac{{(- 1)}^{n + 1}}{(n + 1)!} (n + 1) n (n - 1) \dots 1

= {(- 1)}^{n + 1}

f r o m (A)

P (n + 1) = \frac{Q (n + 1) + (n + 1)}{n + 2}

= \frac{(n + 1) + {(- 1)}^{n + 1}}{n + 2}

Commented by Ar Brandon last updated on 13/Oct/20

Thank You Sir