Prove that: `A_n=3^(n+3)-4^(4n+2)` is divided by `11`

 Prove that: `A_n=3^(n+3)-4^(4n+2)` is divided by `11`

Solution

We will prove by Mathematics Induction:

If `n=0` then `A_n=3^3-4^2=27-16=11` It is true that `A_n` is divided with `11`

Assuming that: `A_n` is divided with `11` for `n=k`        `(1)`

We will prove it is true for `n=k+1` then `A_(k+1)` is divided with `11`

Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`

a. Prove that: `forall x\inR` : `:|acosx+bsinx|\leqsqrt(a^2+b^2)`

b. Find the maximum and minimum of `f(x)=20cosx+21sinx+27`

Solution

a. Prove `|acosx+bsinx|\leqsqrt(a^2+b^2)` `forall x\inR`

Assuming we have: 

`acosx+bsinx=sqrt(a^2+b^2)(a/sqrt(a^2+b^2)cosx+b/sqrt(a^2+b^2)sinx)`

`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0` Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.

`X_1;X_2` are the root of : `X^2-(2cost+3sint)X-11sin^2t=0` 

Find the minimum of `A=X_1^2+X_1X_2+X_2^2`.

Solution

We can see that: `A=X_1^2+X_1X_2+X_2^2`

                                 `=(X_1+X_2)^2-X_1X_2`

Following Vieta's Formulas `X^2-SX+P=0`    `(1)`

Find all Polynomials `P(x)` which is satisfied that: `(x-2010)P(x+67)=xP(x)` (2010 Baltic Way)

 Find all Polynomials `P(x)` which is satisfied that:

`(x-2010)P(x+67)=xP(x)`

(2010 Baltic Way) 

Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`

 a. Prove that: `tan^3x/(1-3tan^2x)=1/8(tan3x-3tanx)`

As we knew: `tan3x=(3tanx-tan^3x)/(1-3tan^2x)`

                                  `=(3tanx-9tan^3x+8tan^3x)/(1-3tan^2x)`

                                  `=(8tan^3x+3tanx(1-3tan^2x))/(1-3tan^2x)`

                                  `=(8tan^3x)/(1-3tan^2x)+3tanx`

Then,             `tan3x-3tanx=(8tan^3x)/(1-3tan^2x)`

Find the limit of `S_n=2/(1.3)+2/(3.5)+2/(5.3)+.......+2/((2n+1)(2n+3))`

 Find the limit of `S_n=2/(1.3)+2/(3.5)+2/(5.3)+.......+2/((2n+1)(2n+3))`

Solution

We can see that general term of this sequence is:

                    `2/((2k+1)(2k+3))=1/(2k+1)-1/(2k+3)`

If `g(x)=f(x)+1-x` , find the value of `g(2020)`

 It is given function `f(x)` determine on `R` , satisfied that:

            `f(1)=1`

            `f(x+5)\geqf(x)+5`

            `f(x+1)\leqf(x)+1`

If `g(x)=f(x)+1-x` , find the value of `g(2020)`   `\forallx,y\inR`

If `(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)` then find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`

1. If `(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)` then find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`

Solution

        From our hypothesis we already had:

                        `(1+2x+3x^2)^10=a_0+a_1x+a_2x^2+.......+a_(20)x^(20)`

        We will find the coefficients of `a_1` ; `a_2` ; `a_3` and `a_20`

        Which meant they are the coefficients of `x` ; `x^2` ; `x^3` and `x^(20)` .

        We can rewrite `(1+2x+3x^2)^10=[1+x(2+3x)]^10`

        We will use the Newton's Formula  

            `(a+b)^n`=`\sum_{i=0}^nC(n,i)``a^(n-i).b^i`

        Then, `[1+x(2+3x)]^10=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2.(2+3x)^2`

                                            `+C(10,3)x^3.(2+3x)^3+....+C(10,10)x^10.(2+3x)^10`

        We observe that: 

            `Coef(x)=2C(10,1)=20` then, `a_1=20`

            `Coef(x^2)=2^2C(10,2)=4.45=180` then, `a_2=180`

           `Coef(x^3)=12C(10,2)+8C(10,3)=45.12+120.8=1500` then `a_3=1500`

            `Coef(x^20)=C(10,10).3^10=3^10` then, `a_20=3^10`     

Prove `(21n+4)/(14n+3)` Is Irreducible For Every Natural Number `n`

Prove `(21n+4)/(14n+3)` Is Irreducible For Every Natural Number `n`. 

Solution

 Solution 01

Denoting the greatest common divisor (GCD) of `a` and `b` `(a,b)` and we will use Euclidean Algorithm Theory.

                `(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1`

It follows that `(21n+4)/(14n+3)` is irreducible. Q.E.D

 Solution 02

 

Vietnam Math Out Standing Student 2012-13

Problem 01: Solve the equation of: `x^(4n)`+`\sqrt{x^{2n}+2012}``=2012`    for all `n\in N`.

Problem 02: It is given `(U_n)` which is determined by: 

                    `U_1=3`

                    `U_(n+1)=1/3(2U_n+3/U_n^2)`;     for all `n\in N`.

                    Let's finding the limit of: `\lim_{n\rightarrow\infty}(u_n)`.

Problem 03: It is given three non-negative real numbers `x,y,z` , Prove that:

                    `1/x+1/y+1/z``>36/(9+x^2y^2+y^2^2+z^2x^2)`.

Problem 04: Find roots positive numbers of equation:

                    `\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z`

Solution

Find all functions `f:R\rightarrow R` such that: `f(x)f(y)=f(xy-1)+xf(y)+yf(x)` `\forall` `x,y` `\inR`

Find all functions `f: R \rightarrow R` Such that:
`f(x)f(y)=f(xy-1)+xf(y)+yf(x)`    `\forall` `x ,y``\inR`

Solution

Replace : `y=0` we will see that: `f(x).f(0)=f(-1)+x.f(0)`

Notice that: `f(0)` not equal to `0` then, `f(x)=x+c`     which `c` is a constant. We replaced to our hypothesis unsatisfactory. Then, `f(0)=0` from that we can see that: `f(-1)=0`.