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

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The problem involves finding the sum of 5a, 25b, 125c, and 625d given the equations involving real numbers a, b, c, and d. By evaluating the polynomial f(x) at specific points and forming a triangular difference table, the fourth differences are found to be constant. Extending the table reveals that f(5) equals -60. Thus, the final result for the sum 5a + 25b + 125c + 625d is -60. The solution was confirmed by multiple participants in the discussion.
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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 $5a+25b+125c+625d$.
 
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[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]
 
Bravo, Opalg! Your answer is correct and I solved the problem using the similar approach as well!(Sun)
 
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