Problem

Question 15

Consider a population that grows according to the recursive rule $P_{n}=P_{n-1}+125$, with initial population $P_{0}=50$.

Then:
\[
\begin{array}{l}
P_{1}= \\
P_{2}=
\end{array}
\]

Find an explicit formula for the population. Your formula should involve $n$ (use lowercase $\mathrm{n}$ )
\[
P_{n}=
\]

Use your explicit formula to find $P_{100}$
\[
P_{100}=
\]

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Answer

\(\boxed{12425}\) is the population at \(n=100\).

Steps

Step 1 :Given the recursive rule \(P_{n}=P_{n-1}+125\) and the initial population \(P_{0}=50\), we can calculate \(P_{1}\) and \(P_{2}\) as follows:

Step 2 :\(P_{1}=P_{0}+125=50+125=175\)

Step 3 :\(P_{2}=P_{1}+125=175+125=300\)

Step 4 :To find an explicit formula for the population, we can observe that the population increases by 125 each time. This is an arithmetic sequence, and the explicit formula for an arithmetic sequence is \(P_{n}=a+(n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Step 5 :In this case, \(a=P_{0}=50\) and \(d=125\). So, the explicit formula for the population is:

Step 6 :\(P_{n}=50+(n-1)125\)

Step 7 :Finally, we can use this explicit formula to find \(P_{100}\):

Step 8 :\(P_{100}=50+(100-1)125=50+99*125=12425\)

Step 9 :\(\boxed{12425}\) is the population at \(n=100\).

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