Suppose $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,.....,n$. Find the value of $P(n+1)$.

Question

QUESTION: Suppose $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,.....,n$. Find the value of $P(n+1)$.

MY APPROACH: At first, this question seemed simple to me.. By putting values of $k$ we can easily get values of $P(0),P(1),....$ and so on. Since, $P(0)=0$ therefore we can say that the constant part of the polynomial is zero. Now,I have assumed the polynomial to be $P(x)=x^n+a_{n-1}x^{n-1}+.....+a_1x+a_0$ But from here things start to mess up.. I am sure this method won't work.. how do I think about it?

Please help me out. Thank you.

score 1 · Answer 1 · answered May 14 '20 at 14:44

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Hint: Consider the polynomial $ Q(x) = (x+1) P(x) - x $. How many values of this polynomial do you know? What is its degree?

answered May 14 '20 at 14:44

Ege Erdil

18,308

Degree is given to be n... And I see $Q(x)$ will be $0$ for all $k=0,1,2,.....,n$.. – Stranger Forever May 14 '20 at 14:51
@StrangerForever The degree of $ Q $ is not $ n $, it's $ n+1 $; and you know the value of $ Q $ at another number... – Ege Erdil May 14 '20 at 14:52

Suppose $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,.....,n$. Find the value of $P(n+1)$.

1 Answers1