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So I have the following equation:

$$c(1)=6$$ $$c(n)=c(n-1)-16$$

-Find the third sequence-

The way I do it: $$c(2)=6(2-1)-16 = -10$$ $$c(3)=-10(3-1)-16 = -36$$ What is it that I do wrong? Why is the correct answer: $$c(3) = -26$$

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  • $\begingroup$ For $c(3)$, $-10-16=-26$ $\endgroup$ – user472341 Sep 18 '17 at 22:34
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    $\begingroup$ $c(n-1)$ isn't equal to $c(n-1) \cdot (n-1)$, right? $\endgroup$ – Stefan4024 Sep 18 '17 at 22:35
  • $\begingroup$ It might be helpful to change $c(n)$ to $c_n$. $\endgroup$ – Math Lover Sep 18 '17 at 22:48
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$c(n)$ is a function. For any $n, c(n)$ has a particular value. You are treating $c$ as a variable above, giving it meaning independent from its meaning as a function.

$c(1) = 6\\ c(2) = c(1) - 16 = 6-16 = -10\\ c(3) = c(2) - 16 = -10 - 16 = -26\\ c(4) = c(3) - 16 = -26 - 16 = -42$

You might be able to reason to a more general equation.

$c(n) = -16n + 22$

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  • $\begingroup$ Hey Doug, thanks for your input. I'm still a little confused.. when you look at the function its $$c(n)=c(n-1)-16$$ Why is it that we don't take the "range" of this function going onward? Like why is it that we don't do: $$c(2)=6(2-1)-16$$ $$or$$ $$c(3)=6(3-1)-16$$ $\endgroup$ – Eli S. Sep 19 '17 at 1:50
  • $\begingroup$ by range I actually meant output.. just to clarify. Fancy with the words. $\endgroup$ – Eli S. Sep 19 '17 at 1:57
  • $\begingroup$ I would just want to understand what part of this function tells us to use the last output as input for the next sequence. Obviously I know its a sequence but I wouldn't know without it being labeled as such. $\endgroup$ – Eli S. Sep 19 '17 at 1:59
  • $\begingroup$ $c(2) = c(2-1) - 16 = c(1)-16$ and $c(3) = c(3-1) - 16 = c(2)-16 = c(1) - 32$ $\endgroup$ – Doug M Sep 19 '17 at 16:36
  • $\begingroup$ $c(n)$ is a function that maps some set of numbers to some other set of numbers. For any number $n$ and $c$ will map that $n$ somewhere else on the number line. $\endgroup$ – Doug M Sep 19 '17 at 16:39
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$$c(2)=c(1)-16=6-16=-10$$

Then:

$$c(3)=c(2)-16=-10-16=-26$$

$c(n-1)$ is the function $c$ evaluated at $n-1$ not some number $c$ multiplied by $n-1$.

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