Sequence and series trivia

[Mohit Pandey]

Hello Hypographites,:eek2:

Here is a question for all --Find out the total of 1 to 100? Give the method or formula used. :lol::confused:

[Erasmus00]

Hello Hypographites,:eek2:

Here is a question for all --Find out the total of 1 to 100? Give the method or formula used. :lol::confused:

 

Divide it into 50 pairs of 101. (1+100, 99+2,3+98... etc). Hence, the sum is 50*101= 5050.

-Will

[Qfwfq]

101*50

 

It can be worked out for the sum of numbers raised to any power but I haven't time to say more right now...

[Mohit Pandey]

Divide it into 50 pairs of 101. (1+100, 99+2,3+98... etc). Hence, the sum is 50*101= 5050.

-Will

How did you find out? Any clue.:lol:

[ronthepon]

You don't know Arithmetic Progression?

 

The formula states:

 

[math]S = {\frac{n}{2}}(a + d(n-1))[/math]

 

If I still fail to light a bulb,

 

S: is the sum you want.

 

a: in the first term of the series

 

d: 'common difference' is the difference between two succesive numbers in the series.

 

n: is the number of terms you wish to add in all.

 

The method Erasmus00 mentioned was first used by Gauss(?), the person who developed the whole concept as a kid.

[Qfwfq]

Of course, in your case Mohit a and d are both 1, so we get n(n + 1)/2. For n = 100 we get 50*101.

 

Erasmus' post makes it easier to understand why, and you can imagine stacking little squares. Draw one square beside a stack of two and then a stack of three etc. You get half of a big square of little squares, including the diagonal. The big square would be n little squares on each side, try working out exactly how many little squares there are using what Erasmus said.

[BrainForce]

Mohit Pandey haven't u read CBSE X maths .I can do these in a second ,i think u are in IX or VIII otherwise u have known it.

actually they form the basic of A.P (what we call arithmetic progression),a very interesting story of gauss is related with it.

 

Now my question on sequences -:

a polygon with 100 sides have how many diagnols?

 

u can use either logic or A.P formulae.

 

 

:D I AM BACK !

[anto]

You don't know Arithmetic Progression?

 

The formula states:

 

[math]S = {\frac{n}{2}}(a + d(n-1))[/math]

 

.

 

I think this formula is wrong ronthepon. The formula should be IMO

 

[math]S = {\frac{n}{2}}(a + a + d(n-1))[/math]

 

because the formula for the sum of an arithemtic series is given by (n/2)(t1 +tn) where tn = (t1 + d(n-1)) so I just think you missed an a

 

I apolagize in advance if im wrong :P

[Mohit Pandey]

Now my question on sequences -:

a polygon with 100 sides have how many diagnols?

 

u can use either logic or A.P formulae.

 

Can anyone help me in solving this question? :confused:

[BrainForce]

See,you havew got to find a pattern ,a pattern which is followed by each polygon ,once u get the pattern nothing can deter u from getting this answer.

 

don't be dependent on formulae,I myself did this question correctly in class VII and used only logic and no formulae.

[CraigD]

a polygon with 100 sides have how many diagnols?

4850.

 

This comes directly from the definition of a diagonal of a polygon as “any line connecting vertexes that is not a side”. The number of lines connecting n vertexes is [math]\frac{n(n-1)}2[/math], because every vertex can form a line with any vertex other than itself (n-1 other vertexes), but a line (a-b) and (b-a) is counted only once (divide by 2). A polygon has the same number of vertexes as sides so the number of diagonals = number of lines – number of sides = [math]\frac{n(n-1)}2-n[/math].

 

Solving for n=100 gives 4850.

[BrainForce]

4850.

 

This comes directly from the definition of a diagonal of a polygon as “any line connecting vertexes that is not a side”. The number of lines connecting n vertexes is [math]\frac{n(n-1)}2[/math], because every vertex can form a line with any vertex other than itself (n-1 other vertexes), but a line (a-b) and (b-a) is counted only once (divide by 2). A polygon has the same number of vertexes as sides so the number of diagonals = number of lines – number of sides = [math]\frac{n(n-1)}2-n[/math].

 

Solving for n=100 gives 4850.

:hyper: that's right!

its good that u formulated it with normal logic,my classmates might do it with formula-n(n-3)/2

[Qfwfq]

I don't see what you mean about your classmates, you don't say why they would use just exactly that formula, on which grounds. Pulled out of a hat? It certainly does follow from Craig's.

don't be dependent on formulae,I myself did this question correctly in class VII and used only logic and no formulae.

But this method does employ a formula!

 

Now, BrainForce, since you seem to be here mainly to pose quizzes, let's see how you do on this one: How many parts is the n-sided polygon divided into by its diagonals?

 

Note: It isn't trivial to work out how many intersection points of more than two diagonals you might have, but you can safely suppose that, for an irregular polygon of any finite number of sides, some adjustment will exist for which there will be none such intersections. An alternative view that I find appealing is that such points just cause some of the parts to have zero area, rather than there being that many less of them.

 

BTW, I saw this question posed elsewhere some time ago and I worked it out, but I don't think I'm gonna do it again too soon... :doh:

[BrainForce]

No! with that i meant to say that i did it without knowing any sequence even n(n-1)/2,just got a pattern with these polygons and got the correct answer.

 

can't we rename this thread to "sequences and series trivia"?????

 

With that question i only wanted to make Mohit aware of such sequences and encourage him to do such questions through some famous books.

[Qfwfq]

just got a pattern with these polygons and got the correct answer.

Well, there are better ways than that!