26. [14] $w, x, y, z$ are real numbers such that
$$
\begin{aligned}
w+x+y+z & =5 \\
2 w+4 x+8 y+16 z & =7 \\
3 w+9 x+27 y+81 z & =11 \\
4 w+16 x+64 y+256 z & =1
\end{aligned}
$$

What is the value of $5 w+25 x+125 y+625 z$ ?
Answer: -60 We note this system of equations is equivalent to evaluating the polynomial (in $a$ ) $P(a)=w a+x a^{2}+y a^{3}+z a^{4}$ at $1,2,3$, and 4 . We know that $P(0)=0, P(1)=5, P(2)=7, P(3)=11$, $P(4)=1$. The finite difference of a polynomial $f$ is $f(n+1)-f(n)$, which is a polynomial with degree one less than the degree of $f$. The second, third, etc finite differences come from applying this operation repeatedly. The fourth finite difference of this polynomial is constant because this is a fourth degree polynomial. Repeatedly applying finite differences, we get
$$
\begin{array}{ccccccccc}
0 & & 5 & & 7 & & 11 & & 1 \\
& 5 & & 2 & & 4 & & -10 & \\
& -3 & & 2 & & -14 & \\
& & & 5 & & -16 & & \\
& & & & -21 & & &
\end{array}
$$
and we see that the fourth finite difference is -21 . We can extend this table, knowing that the fourth finite difference is always -21 , and we find that that $P(5)=-60$. The complete table is