Give P(x) such that P(21) = 17; P(37) = 33.

P(N) = N+51. Find N ?

(NewYork - 1975)

Sorry this question I do for my friend as a reference.

Topic : Zigma on calculator, here I use VINACAL 570ES PLUS II for this problem.

The rule to solve the zigma problem is always formulate the general formula. Last, we must show that the beginning number and the last number is equal ?, after that, all the calculating form is very easy.

Okay now we come to the first example (Violympic 8 - Round 10) :

\(Ex1:\) Let \(a_1,a_2,...,a_n\) are determined by the formula : 

\(a_k=\dfrac{3k^2+3k+1}{\left(k^2+k\right)^3}\forall k\ge1\) Find Sum of : \(1+a_1+a_2+...+a_9?\)

Okay this is an easy problem cause they have given us all the formula and last,first number.

So use calculator : We use SHIFT + log (above ON) :

It appears like \(\sum\limits^{ }_{x=}\left(\right)\), type all the informations into that we have : \(\sum\limits^9_1\left(\dfrac{3x^2+3x+1}{\left(x^2+x\right)^3}\right)\) (Remember always use x to express). Type all the informations then press "=" we have solution \(\dfrac{999}{1000}\).

So the Sum is \(1+\dfrac{999}{1000}=\dfrac{1999}{1000}\).

Easy right ?

Now we come to Example 2 :

Calculate : \(A=\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+99}+\dfrac{1}{50}\)?

So we create the formulate the general formula is \(\dfrac{n}{2}\cdot\left(n+1\right)\). (This formula you must have or else we can't use the calculator). So the general formula is : \(\dfrac{1}{\dfrac{n}{2}\left(n+1\right)}\)

So we continue using calculator to type :D : \(\sum\limits^{99}_{x=2}\left(\dfrac{1}{\dfrac{x}{2}\cdot\left(x+1\right)}\right)\) we have answer : \(\dfrac{49}{50}\)

So the answer is \(\dfrac{49}{50}+\dfrac{1}{50}=1\). Easy right ?

Now we come to harder problem like this :

Example 3 : \(A=\dfrac{2014+\dfrac{2013}{2}+\dfrac{2012}{3}+...+\dfrac{2}{2013}+\dfrac{1}{2014}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}}\)

Find general formula of numerator is \(\dfrac{\left(2015-x\right)}{x}\)

So calculate the numerator we have : \(\sum\limits^{2014}_{x=1}\left(\dfrac{2015-x}{x}\right)=14479,46409\) => Don't worry about the answer, keep it there.

Now we come to calculate the denominator with  general formula is \(\dfrac{1}{x}\)

We have \(\sum\limits^{2014}_{x=1}\left(\dfrac{1}{x}\right)=7,185838258\)

Now the answer is numerator/denominator = \(\dfrac{14479,46409}{7,185838258}=2015\) => Last answer is a nice number.

So that's all, this Zigma's problem I give it to my friend Kayasari Ryuunosuke  and all users in this website.

Thanks :))

Find \(n\left(1010\le n\le2010\right)\) satisfies :

\(a_n=\sqrt{20203+21n}\)

is also a natural number ?

Let a,b are numbers which satisfy \(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\)

Find minimum value of the expression : \(P=a^4+b^4+6a^2b^2\)?

Another good question here :D

Let a,b,c be the lengths of triangle ABC.

The triangle ABC must satisfy what condition that the expression 

\(E=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}\)

has the minimum value ? Find that minimum value.

I have some good math problems in English :D

Number 1. Let a,b be the real numbers satisfy a + b = 2. Prove that \(a^4+b^4\ge a^2+b^2\)

Number 2. Let a,b,c be the positive numbers satisfy ab + bc + ca = 1.

Prove that  : \(\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\le\dfrac{3}{2}\)

Hi Mathulike :0

I know this is a spam question but can I ask a question that : "How can we be the Coordinator in Mathulike and what are standards for us to become a Coordinator ?"

Pls don't remove my question, I'm CTV in page hoc24.vn :)