Math is really tricky!

We note that $f(x)$ is of the following form:

$\begin{aligned} f(x) & = x(x-1)(x-2)(x-3)(x-4) + \color{#D61F06} x^2 + 1 & \small \color{#3D99F6} \text{Note that it is monic and a degree of 5.} \\ f(0) & = 0 + {\color{#D61F06}0^2+1} = 1 & \small \color{#D61F06} \text{See note on how to get }x^2 + 1. \\ f(1) & = 0 + {\color{#D61F06}1^2 + 1} = 2 \\ f(2) & = 0 + {\color{#D61F06}2^2 + 1} = 5 \\ f(3) & = 0 + {\color{#D61F06}3^2 + 1} = 10 \\ f(4) & = 0 + {\color{#D61F06}4^2 + 1} = 17 \\ f(5) & = 120 + 25 + 1 = \boxed{146} \end{aligned}$

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Note: Let $a_0 = 1$ , $a_1=2$ , ... $a_4=17$ . We note that $a_k - a_{k-1} = 2k-1$ . Then we have:

$\begin{aligned} \sum_{k=1}^x a_k - \sum_{\color{#3D99F6}k=1}^{\color{#3D99F6}x} a_{\color{#3D99F6}k-1} & = \sum_{k=1}^x (2k-1) \\ \sum_{k=1}^x a_k - \sum_{\color{#D61F06}k=0}^{\color{#D61F06}x-1} a_{\color{#D61F06}k} & = \sum_{k=1}^x (2k-1) \\ a_x - a_0 & = 2 \times \frac {x(x+1)}2 - x \\ a_x - 1 & = x^2 \\ a_x & = x^2 + 1 \end{aligned}$

A monic polynomial with a degree of $5$ has the form $f(x) = x^5 + Bx^4 + Cx^3 + Dx^2 + Ex + F$ .

For $f(0) = 1$ , $1 = 0^5 + B(0)^4 + C(0)^3 + D(0)^2 + E(0) + F$ , or $F = 1$

For $f(1) = 2$ , $2 = 1^5 + B(1)^4 + C(1)^3 + D(1)^2 + E(1) + F$ , or $B + C + D + E + F = 1$

For $f(2) = 5$ , $5 = 2^5 + B(2)^4 + C(2)^3 + D(2)^2 + E(2) + F$ , or $16B + 8C + 4D + 2E + F = -27$

For $f(3) = 10$ , $10 = 3^5 + B(3)^4 + C(3)^3 + D(3)^2 + E(3) + F$ , or $81B + 27C + 9D + 3E + F = -233$

For $f(4) = 17$ , $17 = 4^5 + B(4)^4 + C(4)^3 + D(4)^2 + E(4) + F$ , or $256B + 64C + 16D + 4E + F = -1007$

Solving the system equations gives $B = -10$ , $C = 35$ , $D = -49$ , $E = 24$ , and $F = 1$ .

Therefore, the polynomial is $f(x) = x^5 - 10x^4 + 35x^3 - 49x^2 + 24x + 1$ , and $f(5) = 5^5 - 10(5)^4 + 35(5)^3 - 49(5)^2 + 24(5) + 1 = \boxed{146}$ .

Relevant wiki: Polynomial Interpolation - Problem Solving

Let $g(x) = f(x)-(x^2+1)$ . Then $g(x)$ is a monic fifth-degree polynomial whose zeroes are $0$ , $1$ , $2$ , $3$ , and $4$ .

Therefore, $g(x) = x(x-1)(x-2)(x-3)(x-4) = f(x)-(x^2+1)$ , so $f(x) = x(x-1)(x-2)(x-3)(x-4)+x^2+1$ . $f(5) = 5(4)(3)(2)(1)+25+1 = \boxed{146}$ .

Because $f(x)=x^2+1$ for $x=0,1,2,3,4$ , we can let $g(x)=f(x)-x^2-1$ , so $0,1,2,3,4$ will be the roots of $g$ . Furthermore, to make $f$ monic, $g$ will need to be monic also, so $g(x)=x(x-1)(x-2)(x-3)(x-4)$ , which let $f(x)=x(x-1)(x-2)(x-3)(x-4)+x^2+1$ , then we have $f(5)=5!+25+1=\boxed{146}$ .

For a polynomial of degree n, the nth set of finite differences will be equal. I remember proving with some algebra 2 students many years ago that not only are they equal, but they are equal to n!*(leading coefficient) as long as the x values are 1 apart. Since the leading coefficient is 1, we just need to make the 5th difference equal to 5!=120. I'll leave a blank at the end of each line until I have the info to fill them in.

$1,2,5,10,17,\_$ Original numbers

$1,3,5,7,\_$ First differences

$2,2,2,\_$ Second

$0,0,\_$ Third

$0,\_$ Fourth

$\_$ Fifth difference

This last blank is the 120.

Backfilling the blanks gives

$120$

$0,120$

$0,0,120$

$2,2,2,122$

$1,3,5,7,129$

$1,2,5,10,17,\boxed{146}$

Not everyone is able to figure out that $f(x)=x^2+1$ for $x=0,1,2,3,4$ . It's OK if you can't think of that. I have a way to overcome that.

Do notice that $f(n+1)-f(n)=2n+1$ for $n=0,1,2,3$ since

$f(1)-f(0)=2-1=\boxed{1},f(2)-f(1)=5-2=\boxed{3},f(3)-f(2)=10-5=\boxed{5},f(4)-f(3)=17-10=\boxed{7}$

$\sum_{n=0}^{k}(f(n+1)-f(n))=\sum_{n=0}^{k}(2n+1)$

$f(1)-f(0)+f(2)-f(1)+f(3)-f(2)+...+f(k)-f(k-1)+f(k+1)-f(k)=2(0)+1+\sum_{n=1}^{k}(2n+1)$

$f(k+1)-f(0)=1+2(\cfrac{k(k+1)}{2})+k$

$f(k+1)-1=1+k(k+1)+k$

$f(k+1)=k^2+2k+2=(k+1)^2+1$

$\therefore f(k)=k^2+1$ for $k=0,1,2,3$

Then, by Factor Theorem , we deduce that $f(x)=x(x-1)(x-2)(x-3)(x-4)+x^2+1$ . Since the question mentions that $f(x)$ is monic polynomial, it follows that the leading coefficient has to be $1$ .

$f(5)=5\times4\times3\times2\times1+5^2+1=\boxed{146}$

f(x) = (x-2)^5 + A(x-2)^4 + B(x-2)^3 + C(x-2)^2 + D(x-2) + 5

Let’s calculate!

Assume six-th point is (5,f5) with fn=f(n), write Lagrange polynomial of degree 5 with 6 points and solve for f5 so that polynomial is monic, i.e. coefficient of x^5 is 1.

Polynomial is sum i fi*[prod (m<>i) (x-xm)/(xi-xm)] for i = 0 to 5 and m in {0,1,2,3,4,5}. Coefficient of x^5 is sum i fi*[prod (m<>i) 1/(xi-xm)]. Thus, it should be 1 for monicity. So, f5 = (1 - sum (i<5) fi/[prod (m<>i) (xi-xm)])*[prod_(m<>5) (xi-xm)].

With the numbers, you get f5 = (1 - (-6+60-300+600-510)/720)*120 = (720+156)/6 = 146