The3Mathketeers

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.

I wrote some computer code, and I found that 11611 is the smallest palindrome divisible by 17. 

In case you are curious, here is the code I wrote for this particular problem. Note I did not finish the code because I already found the smallest palindrome with the incomplete code.

int startNum = 10001;
int numToTest = startNum;
        for (int i = 0; i < 10; ++i){
            for (int j = 0; j < 10; ++j){
                if (numToTest % 17 == 0){
                    System.out.println(numToTest + " is divisible by 17");
                }
                else{
                    System.out.println(numToTest + " is not divisible by 17");
                }
                numToTest = numToTest + 100;
            }
            startNum = startNum + 1010;
            numToTest = startNum;
        }

\(\overline{\rm AC}\) is the hypotenuse of \(\triangle {\rm ACD} \text{ and } \triangle {\rm ABC}\). We can use Pythagorean's Theorem to find that missing length, which will enable us to find the area of the quadrilateral.

Now, we can find the area of the triangles since we have the length of the base and the height of the triangles.

Now, add the areas together to get the area of the whole quadrilateral.

\(A_{\rm ABCD} = A_{\triangle ACD} + A_{\triangle ABC} \\ A_{\rm ABCD} = 4\sqrt{5} + 4\sqrt{2}\)

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.

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

Thanks for the compliment! Your method is equally as valid and potentially quicker than mine. I can claim to have bragging rights as I have been officially featured as "Best Answer"!

I have created a diagram with labeled points so that you can follow along:

In order to solve this problem and find the length \(\rm BC\), we will take advantage of the Chord Intersecting Theorem. This theorem states that, for two intersecting chords within a circle, \({\rm DA} * {\rm AE} = {\rm BA} * {\rm AC}\). In addition, this problem states that \(\overline{\rm BC}\) is tangent to the smaller circle, which means that this segment is perpendicular to the radius of the smaller circle. Since the radius of the larger circle intersects with the chord of the larger circle at a right angle, \(\overline{\rm OD}\) is also a perpendicular bisector of the chord. I have marked this in the diagram. Now, just apply the Chord Intersecting Theorem and solve for for the unknown:

\({\rm DA} * {\rm AE} = {\rm BA} * {\rm AC} \\ 9 * (8 + 8 + 9) = \frac{1}{2}l * \frac{1}{2}l \\ 9 * 25 = \frac{1}{4}l^2 \\ l^2 = 900 \\ l = 30 \text{ or } l = -30\)

Since we are dealing with distances, reject \(l = -30\). Therefore, L = 30.

It is wise to convert modular congruence to equations because we have more experience with equations and know how to manipulate them with ease. \(3n \equiv 1356 \pmod 5 \iff \exists m \in \mathbb{Z} : 3n = 5m + 1356 \).

\(3n = 5m + 1356 \\ n = \frac{5m + 1356}{3}\)

The problem seeks the smallest positive integer, so \(n \geq 1\).

\(n \geq 1 \\  \frac{5m + 1356}{3} \geq 1 \\  5m + 1356 \geq 3 \\  5m \geq -1353 \\ m \geq -\frac{1353}{5} = -270.6\)

We know that \(n = \frac{5m + 1356}{3} = \frac{5}{3}m + 452\). We must select the first value of m that is divisible by 3 so that n will be an integer. Since \(m \geq -270.6\), the first possible value is m = -270. In this case,\( n = \frac{5}{3} * -270 + 452 = -450 + 452 = 2\). In other words, \(n = 2\) is the smallest positive integer n.

Note that this problem only asks for the slope of the perpendicular bisector, which does not require knowing the point of intersection between p and AB. To find the slope, first find the slope between the AB.

\(m_{\overline{\text{AB}}} = \frac{4-7}{3-24} = \frac{-3}{-21} = \frac{1}{7}\)

The relationship between the slope of segment AB and p is the opposite reciprocal. The opposite of 1/7 is -1/7. The reciprocal of -1/7 is -7. Therefore, the slope of p is -7.

\(\)