Messages posted by Hari Shankar

Rewrite the problem by substituting 1 by (a+b+c) as having to prove that

Setting and dividing by each factor on both sides by 2, we can further rewrite as

given x+y+z=1, to prove that which is well known and easy as

Multiplying the above three inequalities concludes the solution

quite obviously (0,1).

If b and c are equal, you can make a as small as you please (as we only need a>b-c)

If a is very large then . But again from triangle inequality a<b+c. Hence the range is (0,1)

Congratulations Vichar!

Well as you have very little time left, I would suggest visiting mathlinks.ro

This is an international forum is most used by students preparing for math olympiads across the world

There is a large bank of questions from contests across the world over several years which you will find very useful

@rishabh.

The angles A,B,C of the triangle are

Hence

So the questions wants us to prove that

But this is true as

Let the number of terms be n. Then the sum of terms of the AP is

Suppose the common ratio of the terms is r, then

Now the sum of terms of the GP can be written as

The typical bracketed term is

We will now prove that this is less than or equal to a+b =

WLOG we may assume that r>1 (otherwise we read the sequence backwards)

So we are to prove that

Hence, each bracketed term is less than or equal to a+b.

Hence the entire sum is less than or equal to thus proving the statement

Since when did we abandon the convention that the remainder is to be less than the divisor?

The correct answer is 9

Hence the given expression equals . Hence the remainder is 1

The solutions above are incorrect

first of all, we are talking sequences really and not functions, and you dont apply derivatives in the conventional sense to them.

Again f(n) being finite has nothing to do with it. Consider f(n) = sin n as n goes to infinity. The sequence is bounded, but it does not converge to any specific value. So saying than sin n = sin (n+1) because as n becomes large n and (n+1) are almost equal would be an error. Clearly if your sequence goes 1,-1,1,-1,... you wouldnt say f(n) = f(n+1)!

That is why proving convergence is important. After that you are justfied in applying this as you are actually applying limits to both sides

That is not the right procedure.

Suppose I give you f(n+1) = f(n)+1, using this procedure you get 1=0. (!!!)

You first need to prove that the sequence is bounded. For this we note that for n>1, we have

Hence the sequence is bounded from below. Note that if a₁= 3, then all members of the sequence equal 3, otherwise they are all greater. We will now assume that

Also, we have

That means the sequence is monotonically decreasing. This sequence is known to converge (in fact to its least upper bound)

Now you can use as done above to obtain the limit as 3

This method was quite popular among the Babylonians to compute the square root of any positive number

You can read more about it at http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method

In fact, we can even give a closed form expression for the general term.

Its easy to see that

The sum of 5 odd numbers is also odd. So that cannot be 50,. Pls chk the question

http://www.goiit.com/posts/list/algebra-question-904371.htm

even shorter, ab+bc+ca =23 and abc=-15.

Hence

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