A company charges 8.00 to dig a hole one foot deep. The second foot is 5.40, the third $4.32, etc. (Note that the cost of each subsequent foot is 20% less than the previous). The question asks what it will cost to dig a hole 5 feet deep. I can arrive at the solution alright by making a chart - but how can I write an expression and teach it to my students?
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1@Kaster I think a suitable runner-up to actual knowledge is the ability to ask for help, rather than the steadfast belief that your knowledge is sufficient and/or that the topic is irrelevant. Surely this is a worthy quality to teach students, as well. – Emily Feb 21 '13 at 03:32
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4So for $40, I can have them keep digging forever? Cool! – hmakholm left over Monica Feb 21 '13 at 03:32
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Chart is good. I would urge that you not just fill it out with numbers out of a calculator. So do what you presumably did, labels first foot, second foot, and so on. In first, of course you put $8$. For second, put $\frac{80}{100}\cdot 8$, like that. For third, put $\frac{80}{100}\frac{80}{100}\cdot 8$, and maybe $\left(\frac{80}{100}\right)^2\cdot 8$. And so on. Whether you introduce formula for sum of GP is a matter of grade. If you don't, maybe provide a link for the interested. After all this, you can use the calculator to give conventional "evaluations" of the various terms. – André Nicolas Feb 21 '13 at 03:34
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1For sum, it is probably useful to introduce at least the expression $a+aq+aq^2+aq^3+aq^4$, where $a$ serves as abbreviation for $8$, and $q$ for $80/100$. – André Nicolas Feb 21 '13 at 03:43
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80% of $8.00 is $6.40 not $5.40 --- does the first foot not fit the pattern, or is there a mistake in the question? – Glen_b Feb 21 '13 at 05:45
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As anon mentions, the quantity you are looking at has the form $$a+ar+ar^2+ar^3+ar^4$$ where $a=8$ and $r=\frac{80}{100}$. This type of series is called a geometric series, and there is a comprehensive Wikipedia page which provides an enormous amount of details. In particular, you are interested in a general formula for this series. Both the formula, and the proof for why it is true can be found in the formula section of the Wikipedia page. I have copied the relevant portion below.
For $r\neq 1$, the sum of the first $n$ terms of a geometric series is: $$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = a\cdot \frac{1-r^{n}}{1-r}.$$ here $a$ is the first term of the series, and $r$ is the common ratio. We can derive this formula as follows: $$\text{Let }s = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}.$$ $$\text{Then }rs = ar + ar^2 + ar^3 + ar^4 + \cdots + ar^{n}$$ $$\text{Then }s - rs = a-ar^{n}$$ $$\text{Then }s(1-r) = a(1-r^{n}),$$ $$\text{ so }s = a \frac{1-r^{n}}{1-r}.$$
Eric Naslund
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