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Knowing that $y(k+1)=1.02y(k)$ for every $k=1,2,...$
And $y(0)=0$.

And I'm asked to draw the graph of $y(k)$ as a function of $k$ for $k=1,2,...,100.$

I've thought of calculating some of $y(1),y(2),...$, but that seems so inefficient.

So I've started trying to find a pattern:
$y(0)=100. $
$y(1)=1.02y(0)$.
$y(2)=1.02^2 y(0)$.
$y(3)=1.02^3*100$.
.
.
$y(k)=1.02^k*100$.
Now I could just plug in a couple of values and draw a line to connect them to approximately draw my graph.

I would appreciate any feedback on my solution and if there's any other more better ways.

Pwaol
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    Possible mistake. According to calculation, condition should be the following: $y(k+1)=1.02 y(k)$, for every $k=0,1,2,...$, $y(0)=100$. Because of discrete set of possible $k$ it is correct to draw a graph in form of set of points $(k,y(k))$ without any connections. All these points will lie at one exponent function $y=100\cdot 1.02^x$ graph, so you can plot additionally this exponent graph with dotted line to show this fact. – Ivan Kaznacheyeu Nov 01 '21 at 10:37
  • If I am understanding your idea, that you can just plot the two end points and draw the line segment connecting them, then that is wrong. This is an exponential function, not a line. It curves upward as $k$ increases. – Paul Sinclair Nov 01 '21 at 19:02

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