a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5$ 
Let $P(x)$ be a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5.$ Determine the value of $P(6)$.

Let $P(x) = ax^4 + bx^3 + cx^2 + dx + e$. For $n=1,2,3,4,5$ I have plugged it into this polynomial and got the following —
$$P(1) = a+b+c+d+e = 120$$
$$P(2) = 16a + 8b + 4c + 2d + e = 60$$
$$P(3) = 81a + 27b + 9c + 3d + e = 40$$
$$...$$
And what the problem asks for is $$P(6) = 1296a + 216b + 36c + 6d + e .$$
However,  I'm not sure if all this is helping me very much. So noticing that $2P(2) = P(1) = 3P(3)$ (which is also equal to $4P(4), 5P(5)...$) From solving simultaneous equations I got that $31a + 15b + 7c + 3d + e=0$ and similarly $211a + 65b + 19c + 5d + e=0$, but they seem rather useless at this point. 
 A: 
So noticing that $2P(2) = P(1) = 3P(3)$ (which is also equal to $4P(4), 5P(5)...$)

You are on the right track. $$x P(x) - 120$$ is a polynomial of degree (at most) 5, and has zeros at $x= 1, 2, 3, 4, 5$, therefore
$$
x P(x) - 120 = c(x-1)\cdots (x-5)
$$
for some constant $c$, which can be determined by substituting $x = 0$.
A: Hint:
$$P(n)=n(n-1)(n-2)(n-3)(n-4)\sum_{r=0}^4\dfrac{a_r}{n-r}$$
A: Hint: We can write $P(x)$ in the form of
$$P(x) =c_1(x-2)(x-3)(x-4)(x-5)\ +\ c_2(x-1)(x-3)(x-4)(x-5)\ +\dots+\ c_5(x-1)(x-2)(x-3)(x-4)$$
A: The first $5$ values uniquely determine the polynomial of degree $4$. By Lagrange interpolation we have
$$
P(x)=x^4−15x^3+85x^2−225x+274.
$$
We obtain $P(6)=40$.
A: The previous solutions calculate the polynomial, and then deduce the value of $P(6)$.
This was done correctly of course (the solution by @Martin R is particularly elegant).
However, it is possible to calculate the value of $P(6)$ directly, by calculating the finite differences and by using the fact that the 5th-order difference is equal to $0$ for polynomials of degree $4$.
$$
\begin{array}{c|rrrrrr}
x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
P(x) & 120 & 60  & 40  & 30  & 24 & \color{red}{40} \\
\Delta_1 &       & -60 & -20 & -10 & -6 & \color{red}{16} \\
\Delta_2 &       &     &  40 &  10 &  4 & \color{red}{22} \\
\Delta_3 &       &     &     & -30 &  6 & \color{red}{18} \\
\Delta_4 &       &     &     &     & 24 & \color{red}{24} \\
\Delta_5 &       &     &     &     &    & \color{red}{0}
\end{array}
$$
In the above array, the values in black are derived from the input data, and the values in red are calculated, starting from the value $\Delta_5=0$.
And the solution is $P(6) = 40$.
