Find the sum of 5a, 25b, 125c and 625d

[anemone]

Given $a,\,b,\,c,\,d$ are real numbers such that

$a+b+c+d=5$

$2a+4b+8c+16d=7$

$3a+9b+27c+81d=11$

$4a+16b+64c+256d=1$

Evaluate $5a+25b+125c+625d$.

 

[Opalg]

[sp]Let $f(x) = ax + bx^2 + cx^3 + dx^4$. Then we know that $f(0) = 0$, $f(1) = 5$, $f(2) = 7$, $f(3) = 11$ and $f(4) = 1.$ Suppose we form the repeated differences between these values, in a triangular table like this (where each element, apart from those in the top row, is the difference between the two elements above it):

$$\begin{array}{ccccccccc} 0&&5&&7&&11&&1 \\ &5&&2&&4&&-10& \\ &&-3&&2&&-14&& \\ &&&5&&-16&&& \\ &&&&-21&&&& \end{array}$$

Since $f(x)$ is a fourth-degree polynomial, its fourth differences must be constant, so we can extend the table, from the bottom row upwards, knowing that the elements in the bottom row must all be $-21$. The extended table looks like

$$\begin{array}{ccccccccccc} 0&&5&&7&&11&&1&& \color{red}{-60} \\ &5&&2&&4&&-10&& \color{red}{-61}& \\ &&-3&&2&&-14&& \color{red}{-51}&& \\ &&&5&&-16&& \color{red}{-37}&&& \\ &&&&-21&& \color{red}{-21}&&&& \end{array}$$

By the time we get back up to the top row, we see that $f(5) = -60$.[/sp]

 

[anemone]

Bravo, Opalg! Your answer is correct and I solved the problem using the similar approach as well!(Sun)