Question: Why divide by {some integer} .5 at 8:36?

[The Blade Dancer]

Why divide by {some integer} .5 at 8:36? I get how if you calculate it, the median will be {some integer} .5, but why should we divide by it?

[debbie]

@The-Blade-Dancer I'm very sorry for the delay, but I will get to this tomorrow morning

[debbie]

@The-Blade-Dancer I'm looking for the place in the video where you are mentioning about the dividing by {some integer}.5.... I can't seem to find Prof. Loh saying this exact line... He does talk about multiplying \(30\) with {some integer}.5, and maybe I'll backtrack a bit to review what he said.

In the video at around 8:36, Prof. Loh is showing us an expression for the sum of \(30\) consecutive numbers.

"Whenever you have an even number of consecutive integers, then the evenness and oddness of the first and last are different."

You can see that this is true by listing out some sample consecutive sequences with an even number of terms, like this:

\(\textcolor{red}{1 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{2} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{red}{3 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{4} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{red}{5 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{6} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{red}{7 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{8} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{red}{9 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{10} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{red}{11 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{12} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{red}{13 } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \textcolor{blue}{14} \text{ } \text{ } \text{ }\)

\( \textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ } \textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ }\textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ }\textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ }\textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ } \textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ }\textcolor{red}{\text{odd }} \text{ } \text{ } \text{ } \textcolor{blue}{\text{even}} \text{ } \text{ } \text{ }\)

Therefore, the sum of the \(\textcolor{red}{\text{first}}\) and \(\textcolor{blue}{\text{last}}\) will always be odd, and the average will always be a whole number plus \(0.5.\)

(Note: You mentioned the median, and it happens to be true that for a sequence where the terms are equally spaced apart (found by adding the same number to a term to get the next term), the median and the average/mean is the same. In general, though, for a any old sequence, we can't interchange the words median and mean. The median is the middle number if you arrange the numbers from smallest to largest. For example, for the first eleven terms of the Fibonacci sequence, the median is \(\textcolor{red}{8},\) but the mean is \( \frac{\textcolor{red}{1 + 89}}{2} = \textcolor{red}{45}.)\)

\( \textcolor{red}{1 \text{ }\text{ }\text{ }\text{ } \text{ }\text{ }\text{ } 1 \text{ }\text{ }\text{ }\text{ } \text{ } \text{ }2 \text{ }\text{ }\text{ }\text{ }\text{ } \text{ } 3 \text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \text{ }\text{ }5 \text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \text{ }\text{ }8 \text{ }\text{ }\text{ }\text{ }\text{ } \text{ } \text{ }\text{ }13 \text{ }\text{ }\text{ }\text{ } \text{ }\text{ }\text{ } 21 \text{ }\text{ }\text{ }\text{ } \text{ } \text{ }\text{ }34 \text{ }\text{ }\text{ }\text{ } \text{ } \text{ }\text{ }\text{ }\text{ }55 \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } 89 }\)

"What happens if I multiply \(30\) by something ending in \(0.5?\) The easiest way to think about this is to think about what \(xxx.5\) means as a fraction."

Prof. Loh's next step is to replace the average by the fraction \(\frac{\text{odd}}{2},\) which gives us

$$ 30 \times \frac{\text{odd}}{2} $$

$$ 15 \times \text{ odd} $$

It is from here that we can start to look at the answer choices to decide which of them could be the possible sum of the \(30\) consecutive integers. It's the one that is the multiple of \(15!\)

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