The3Mathketeers

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 #5
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The alternating sum or difference of binomial coefficients is 0, which is a fact I already knew. However, I was curious why this happens every time, so I decided to do some summation algebra to prove the result. If you are confused at any step, I can explain some steps more, but I added light commentary where I saw fit.

 

\(\begin{align*} \sum_{i = 0}^{n}(-1)^i \binom{n}{i} &= (-1)^0\binom{n}{0} + (-1)^n\binom{n}{n} + \sum_{i = 1}^{n - 1}(-1)^i\binom{n}{i} \\ &= 1 + (-1)^n + \sum_{i = 1}^{n - 1}(-1)^i\left[\binom{n-1}{i-1} + \binom{n - 1}{i}\right] \text{by Pascal's Rule} \\ &= 1 + (-1)^n + \sum_{i = 1}^{n - 1} (-1)^i\binom{n-1}{i-1} + (-1)^i\binom{n-1}{i} \end{align*}\)

 

You might wonder why I did all of this. After all, it looks like I am making the expression more complicated. However, I will use the power of a telescoping sum to simplify this monstrosity further. I apologize that I am not experienced enough with LaTeX to make this look more visually appealing.

 

\(\sum_{i = 1}^{n - 1} (-1)^i\binom{n-1}{i-1} + (-1)^i\binom{n-1}{i} \\ \begin{align*} i = 1:& (-1)\binom{n-1}{0} &+(-1)\binom{n-1}{1} & \\ i = 2:& &+(1)\binom{n-1}{1} & + (1)\binom{n-1}{2} \\ i = 3:&&& + (-1)\binom{n-1}{2} + (-1)\binom{n-1}{3}\\ \cdots \end{align*}\)

 

This makes it clear that the summation will collapse to the first and last term of the summation.

 

\(\begin{align*} \sum_{i = 1}^{n - 1} (-1)^i\binom{n-1}{i-1} + (-1)^i\binom{n-1}{i} &= (-1)^1\binom{n-1}{0} + (-1)^{n-1}\binom{n-1}{n-1} \\ &= -1 + (-1)^{n - 1} \end{align*}\)

 

Now, we put this information together and see to what the original sum evaluates.

 

\(\begin{align*} \sum_{i = 0}^{n}(-1)^i \binom{n}{i} &= 1 + (-1)^n -1 + (-1)^{n -1} \\ &= (-1)^n + (-1)^{n-1} \\ &= (-1)^n(1 - 1) \\ &= 0 \end{align*}\)

 

This concludes the proof and proves that the alternating sum or difference of binomial coefficients is 0.

16.08.2023
 #2
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I have created a diagram, and I added a few points for convenience. 

 

Notice that \(\overline{\rm BC} \perp \overleftrightarrow{\rm CD} \text{ and } \overline{\rm AD} \perp \overleftrightarrow{\rm CD}\) because of the Tangent Perpendicular to Radius Theroem. Since both segments are perpendicular to the same line, \(\overline{\rm AD} \parallel \overline{\rm BC}\). Quadrilateral ACBD has a pair of parallel sides and a pair of non-parallel sides, so this quadrilateral is classified as a trapezoid. The area formula for a trapezoid is \(A_{\text{trapezoid}} = \frac{1}{2}\left(b_1 + b_2\right)h \) where b1 and b2 are the lengths of each parallel side and h is the perpendicular distance between b1 and b2 of the trapezoid. I have constructed the diagram such that \(\overline{\rm CE} \parallel \overline{\rm AB}\). Now, pay close attention to \(\triangle {\rm EDC}\). This is a right triangle, so we can use Pythagorean's Theorem to find \(\rm CD\), which is the height of the trapezoid.

 

\({\rm CD}^2 + {\rm ED}^2 = {\rm EC}^2 \\ {\rm CD}^2 + (2 + 1)^2 = 6^2 \\ {\rm CD}^2 = 36 - 9 \\ {\rm CD}^2 = 27 \\ {\rm CD} = 3\sqrt{3} \text{ or } {\rm CD} = -3\sqrt{3}\)

 

Since we are dealing with lengths, we should reject \({\rm CD} = -3\sqrt{3}\). Now, we have all the ingredients to find the area of the trapezoid ACBD.

 

\(A_{\text{trapezoid}} = \frac{1}{2}\left(b_1 + b_2\right)h \\ A_{\text{trapezoid}} = \frac{1}{2}(2 + 1)*3\sqrt{3} \\ A_{\text{trapezoid}} = \frac{9\sqrt{3}}{2} \approx 7.7942 \)

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15.08.2023