Un 1 UviSiDle Dy (28) Let 01 =5,and let 0n+1 a2 Prove that the last n digits of an are the same as the last n digits of an+1. (29) Prove that for any positive integ...

Proof In Exercises $99-102,$ prove the property for all
integers $r$ and $n$ where $0 \leq r \leq n .$
The sum of the numbers in the $n$ th row of Pascal's
Triangle is $2^{n}$ .

Okay, The first thing we know from this question is that we're trying to prove the triangular formula and princes ample swan divide by two. So we know that something numbers up, we're gonna be having multiplication and division. So looking at a visual proof, we know that if we've got half a rectangle with dots five toll and six wide, we have half of 30 dots, which is gonna be 15th for t a five. So if we have one post 31 post to Post Reports and this is gonna be an prophecies samples one divided by two. So if you were to look at the proof using algebra, we would get chu tea of on equals. And the reason why is that? We just established. But t even pushed even equals one plus two plus three plus and minus one plus and it continues going on forward. It's a proof using algebra. So this means that t of an is gonna be an times on plus one divided by two. So we can also look at this using signal notation. If you look at this using sigma notation, you would see this was the algebraic we now get the signal. A signal notation. I was, um, As you can see, this looks very similar to what we have earlier. It's just that we have signal detection here so we can see that we have any number of copies of N plus One, which means the eye does not appear in the formula. So all the terms are equivalent to each other, giving us t of an equals and parentheses and plus one divided by two.

We're asked you strong induction to prove that when a simple polygon p with consecutive Verdecia be one through the end is triangulated into end minus two triangles. The end minus two triangles could be numbered 12 up their end minus two. So that V I is going to get Vertex of triangle I for I from one end minus two. So let P N b the statement that the end minus two triangles and the number 12 up Teoh and minus two such that the eyes of Vertex of triangle I for i e going one up two and minus two. So clearly we can't triangulate simple polygon with anything less than three courtesies. So for of this statement for the basis statement first So in this case, n equals three So a convex polygon p the three birdies scenes is a triangle and thus is the triangulation of P name the entire triangle of the one. Then we have that triangles have been labeled such that the one is a vertex of triangle one. Sorry is trying one so have shown that p three is true and to use strong induction Let's suppose that p one starting at one p three up through PKR. True for some K greater than or equal three. We want to prove that P K plus one is true. So let P K plus one a convex Holly Gagne with Vergis sees a one B to weigh up to V K plus one. These are actually consecutive Verdecia section C. They're connected by edges each the truth and let Devia diagonal its present a triangulation of peed and minus one triangles. And we had this. Dagnall has to exist. Or else there will be more than an minus two triangles, I mean and minus one triangles. And let's assume that the Dagnall D is between Vergis ease V one and V I where we have that I is gonna be anywhere from three decay. It's any non adjacent vertex. And of course, if not in this relay Bolivar theses of the polygon. So we have that convicts Holly Gagne keep prime with courtesies. B one B two up to he I You're consecutive courtesies waas a triangulation with I minus two triangles and the I minus two triangles can be labeled 12 up through I minus two, such that VJs a vertex of Triangle J. This is because the statement t J p J p i is true since I z going to be less than K plus one. And likewise we have the convicts Polygon p double prime with the consecutive Vergis is V I plus one up to the K plus u K plus one. The one has a triangulation with. So the number of burgesses in this case is going to be K plus one minus I and then from that will subtract two to get the number of triangles in these triangles. And he labelled well as I I plus one up to In this case we get k plus one minus I minus two. So it's k minus one minus. I such that we have the J is going to be a vertex of Triangle J. This is because the statement p k plus warn minus I it's true again. I is going to be less than faithless one that are equal to one. And so this form the triangulation of P such that the end minus two triangles. I'm sorry, the K minus two triangles, You should really be k minus one. Sorry came on this one. Triangles we have can be numbered 12 all the way up to K minus one. And so that her text VJ z vertex of Triangle J. And so this shows that P K plus one is true in the US by strong induction. It follows that PN is true girl and greater than or equal to.

So in the problem we're gonna have to do with possible name or triangle can grow into the given triangle and state the congruence conjecture. If you can't show any triangles to be congruent for the information right, can't be determined and explain why. So Triangle Man, Triangle man Is equal and can grow into Triangle one. So for this problem it can't be determined. So can B Germond? That's wrong spelling, but it's okay, so that it can't be determined because the parts do not correspond, so the car parts parts do not correspond. So that's why you can't be determined.