Does pattern always true?

Define $q(x)=(x+1)p(x)-x$

Note that for all n=0, 1, 2, 3, 4,

$p(n)=\frac{n}{n+1}$

Therefore, $q(n)=(n+1)\frac{n}{n+1}-n=n-n=0$

So, the polynomial q(x) has roots 0, 1, 2, 3, 4

As a result, we can rewrite q(x) into the form $Kx(x-1)(x-2)(x-3)(x-4)$

where K is a real constant.

Recall, by definition, that $q(x)=(x+1)p(x)-x$

Therefore, $(x+1)p(x)-x=Kx(x-1)(x-2)(x-3)(x-4)$

To solve for K, simply put x=-1 to the identity.

Now we have $(-1+1)p(-1)-(-1)=K(-1)(-1-1)(-1-2)(-1-3)(-1-4)$

Simplify, we have $1=-120K$

$K=\frac{-1}{120}$

Thus, we have $q(x)=\frac{-1}{120}x(x-1)(x-2)(x-3)(x-4)$

Put x=5 into the expression, we have $q(5)=\frac{-1}{120}(5)(5-1)(5-2)(5-3)(5-4)=-1$

By the definition of q(x), $q(5)=6p(5)-5$

Therefore, $p(5)=\frac{q(5)+5}{6}=\frac{-1+5}{6}=\frac{2}{3}=\boxed{0.667}$

For those of you who say the answer is $\frac{5}{6}$ , you are wrong because p(x) is a polynomial, not a sequence!

The question is come from Hong Kong Advanced Level Examination 2006 Pure Mathematics Paper 1 question 3

My friend in Hong Kong did't know how to solve this and sent the picture to me. It is pretty cool so I cancel out some part of the question and post it here.

A standard way of solving this type of problem is as follows:

\begin{aligned} p(x) & = \sum_{k=0}^4 a_k \prod_{\substack{j=0 \\ j \ne k}}^4 (x-j) & \small \color{#3D99F6} \text{where }a_k \text{ is a coefficient.} \end{aligned} $\quad \ \ = \underbrace{a_0(x-1)(x-2)(x-3)(x-4)}_{\color{#3D99F6}\text{No }(x-0)} + \underbrace{a_1x(x-2)(x-3)(x-4)}_{\color{#3D99F6}\text{No }(x-1)} + \underbrace{a_2x(x-1)(x-3)(x-4)}_{\color{#3D99F6}\text{No }(x-2)} + \underbrace{a_3x(x-1)(x-2)(x-4)}_{\color{#3D99F6}\text{No }(x-3)} + \underbrace{a_4x(x-1)(x-2)(x-3)}_{\color{#3D99F6}\text{No }(x-4)}$ $\begin{aligned} p(0) & = a_0(-1)(-2)(-3)(-4) = 0 & \small \color{#3D99F6} \implies a_0 = 0 \\ p(1) & = a_1(1)(-1)(-2)(-3) = \frac 12 & \small \color{#3D99F6} \implies a_1 = -\frac 1{12} \\ p(2) & = a_2(2)(1)(-1)(-2) = \frac 23 & \small \color{#3D99F6} \implies a_2 = \frac 16 \\ p(3) & = a_3(3)(2)(1)(-1) = \frac 34 & \small \color{#3D99F6} \implies a_2 = -\frac 18 \\ p(4) & = a_4(4)(3)(2)(1) = \frac 45 & \small \color{#3D99F6} \implies a_2 = -\frac 1{30} \end{aligned}$

Therefore, we have:

$\begin{aligned} p(x) & = - \frac {x(x-2)(x-3)(x-4)}{12} + \frac {x(x-1)(x-3)(x-4)}6 - \frac {x(x-1)(x-2)(x-4)}8 + \frac {x(x-1)(x-2)(x-3)}{30} \\ \implies p(5) & = - \frac {5(3)(2)(1)}{12} + \frac {5(4)(2)(1)}6 - \frac {5(4)(3)(1)}8 + \frac {5(4)(3)(2)}{30} \\ & = - \frac 52 + \frac {20}3 - \frac {15}2 + 4 \\ & = \frac 23 \approx \boxed{0.667} \end{aligned}$

A polynomial of degree 4 with real coefficients has the equation $y = ax^4 + bx^3 + cx^2 + dx + e$ .

When $p(0) = 0$ , $0 = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e$ , so $e = 0$ , and $y = ax^4 + bx^3 + cx^2 + dx$ .

When $p(1) = \frac{1}{2}$ , $\frac{1}{2} = a(1)^4 + b(1)^3 + c(1)^2 + d(1)$ or $2a + 2b + 2c + 2d = 1$ .

When $p(2) = \frac{2}{3}$ , $\frac{2}{3} = a(2)^4 + b(2)^3 + c(2)^2 + d(2)$ or $24a + 12b + 6c + 3d = 1$ .

When $p(3) = \frac{3}{4}$ , $\frac{3}{4} = a(3)^4 + b(3)^3 + c(3)^2 + d(3)$ or $108a + 36b + 12c + 4d = 1$ .

When $p(4) = \frac{4}{5}$ , $\frac{4}{5} = a(4)^4 + b(4)^3 + c(4)^2 + d(4)$ or $320a + 80b + 20c + 5d = 1$ .

Solving this system of equations gives $a = -\frac{1}{120}$ , $b = \frac{11}{120}$ , $c = -\frac{23}{60}$ , and $d = \frac{4}{5}$ .

So $p(x) = -\frac{1}{120}x^4 + \frac{11}{120}x^3 - \frac{23}{60}x^2 + \frac{4}{5}x$ , which means $p(5) = -\frac{1}{120}(5)^4 + \frac{11}{120}(5)^3 - \frac{23}{60}(5)^2 + \frac{4}{5}(5) = \frac{2}{3} \approx 0.667$ .

Click here to view the solution.