If $P$ is a fifth degree polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0, 1, 2, 3, 4$ and $5$, then find $P(6)$.

Question

So I let $P(x)=ax^5+bx^4+cx^3+dx^2+ex+f$, and obviously $f=0$. Then,

$2^5a+2^4b+2^3c+2^2d+2e={1\over 2}$

$3^5a+3^4b+3^3c+3^2d+3e={2\over 3}$

$4^5a+4^4b+4^3c+4^2d+4e={3\over 4}$

$5^5a+5^4b+5^3c+5^2d+5e={4\over 5}$

and we are looking for $6^5a+6^4b+6^3c+6^2d+6e=$

Then I got stuck... Could anyone at least give me some hints or methods to continue?

Try interpolation (Lagrange, Newton). Also notice k/(k+1)=1-1/(k+1). — Divide1918, Dec 17 '20 at 08:46
Also: https://math.stackexchange.com/q/636919, https://math.stackexchange.com/q/389600 — Martin R, Dec 17 '20 at 08:53

score 6 · Accepted Answer · answered Dec 17 '20 at 08:52

Hint: Note that $Q(x) = (x+1)P(x)-x$ is a polynomial of degree at most $6$ which satisfies $Q(k) = 0$ for $k = 0,1,2,3,4,5$. Hence, $Q(x) = Cx(x-1)(x-2)(x-3)(x-4)(x-5)$ for some constant $C$. To solve for $C$, try to figure out what $Q(-1)$ must be. Then, finding $P(x)$, and thus, $P(6)$ will be easy.

If $P$ is a fifth degree polynomial such that $P(k)=\frac{k}{k+1}$ for $k=0, 1, 2, 3, 4$ and $5$, then find $P(6)$.

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