heureka

​Find x. Round to the nearest tenth if necessary.

\(\begin{array}{|rcll|} \hline x\cdot (x+1) &=& 10 \cdot 3 \\ x^2+x &=& 30 \\ x^2+x-30 &=& 0 \\ (x-5)(x+6) &=& 0 \\\\ x_1 &=& 5 \\ x_2 &=& -6 \qquad \text{no solution, because }x \lt 0 \\ \hline \end{array} \)

d. 5

Determine all positive integers
\(k \le 2000 \)  
for which
\(x^4+k\)
can be factored into two distinct trinomial factors with integer coefficients.

\(\text{All four roots of $x^4+k$ are complex numbers, because $k$ is a positive integer} \\ \text{The complex numbers are conjugation in pairs :} \)

\(\begin{array}{|rcll|} \hline x^4+k &=& (x-x_1)(x-x_2)(x-x_3)(x-x_4) \quad | \quad x_1,x_2,x_3,x_4 \in C \\ && x_1 = a+bi \\ && x_2 = a-bi \\ && x_3 = u+vi \\ && x_4 = u-vi \\ x^4+k &=& \Big(x-(a+bi)\Big) \Big(x-(a-bi)\Big) \Big(x-(u+vi)\Big) \Big(x-(u-vi)\Big) \\ &=& \Big((x-a)-bi\Big) \Big((x-a)+bi)\Big) \Big((x-u)-vi)\Big) \Big((x-u)+vi)\Big) \\ &=& \Big((x-a)^2-b^2i^2\Big) \Big((x-u)^2-v^2i^2)\Big) \quad | \quad i^2=-1\\ &=& \Big((x-a)^2+b^2\Big) \Big((x-u)^2+v^2)\Big) \\ &=& \Big(x^2-2ax+(a^2+b^2)\Big) \Big(x^2-2ux +(u^2+v^2)\Big) \quad | \quad \text{trinomial factors}\\ &=& x^4-2ux^3+(u^2+v^2)x^2-2ax^3+4aux^2-2ax(u^2+v^2) \\ && +(a^2+b^2)x^2-2u(a^2+b^2)x+(a^2+b^2)(u^2+v^2) \\ &=& x^4 -x^3\underbrace{(2u+2a)}_{=0} +x^2\underbrace{(u^2+v^2+a^2+b^2+4au)}_{=0} \\ && -x\underbrace{\Big(2a(u^2+v^2)+2u(a^2+b^2)\Big)}_{=0} +\underbrace{(a^2+b^2)(u^2+v^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 1. & 2u+2a &=& 0 \quad & | \quad : 2 \\ & u+a &=& 0 \\ & \mathbf{u} &\mathbf{=}& \mathbf{-a} \quad & | \quad \text{or} \quad \mathbf{u^2=a^2} \\ \hline \end{array}\\ &=& x^4 +x^2\underbrace{(a^2+v^2+a^2+b^2-4a^2)}_{=0} \\ && -x\underbrace{\Big(2a(a^2+v^2)-2a(a^2+b^2)\Big)}_{=0} +\underbrace{(a^2+b^2)(a^2+v^2)}_{=k} \\ &=& x^4 +x^2\underbrace{(v^2+b^2-2a^2)}_{=0} -x\underbrace{(2av^2-2ab^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+v^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 2. & 2av^2-2ab^2 &=& 0 \quad & | \quad : 2a \\ & v^2-b^2 &=& 0 \\ & \mathbf{v^2 } &\mathbf{=}& \mathbf{b^2 } \\ \hline \end{array}\\ &=& x^4 +x^2\underbrace{(b^2+b^2-2a^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+b^2)}_{=k} \\ &=& x^4 +x^2\underbrace{(2b^2-2a^2)}_{=0} +\underbrace{(a^2+b^2)(a^2+b^2)}_{=k} \\ && \begin{array}{|lrcll|} \hline 3. & 2b^2-2a^2 &=& 0 \quad & | \quad : 2 \\ & b^2-a^2 &=& 0 \\ & \mathbf{b^2 } &\mathbf{=}& \mathbf{a^2 } \\ \hline \end{array}\\ &=& x^4 +\underbrace{(a^2+a^2)(a^2+a^2)}_{=k} \\ &=& x^4 +\underbrace{(2a^2)(2a^2)}_{=k} \\ &=& x^4 +\underbrace{4a^4}_{=k} \\ \hline && \huge{k=4a^4} \\ \hline \end{array}\)

\(\text{trinomial factors:}\)

\(\begin{array}{|rcll|} \hline && \Big(x^2-2ax+(a^2+b^2)\Big) \Big(x^2-2ux +(u^2+v^2)\Big) \\ && \boxed{ b^2 = a^2, \quad u^2 = a^2, \quad v^2=b^2=a^2, \quad u = -a } \\ &=& ( x^2-2ax+ 2a^2)(x^2+2ax + 2a^2) \\ \hline \end{array} \)

Solutions:

\(\begin{array}{|r|r|c|c|} \hline a \in N &k= 4a^4 & k \le 2000 & ( x^2-2ax+ 2a^2)(x^2+2ax + 2a^2) \\ \hline 1 & 4 & \checkmark & ( x^2-2x+ 2)(x^2+2x + 2) \\ 2 & 64 & \checkmark & ( x^2-4x+ 8)(x^2+4x + 8) \\ 3 & 324 & \checkmark & ( x^2-6x+ 18)(x^2+6x + 18) \\ 4 & 1024 & \checkmark & ( x^2-8x+ 32)(x^2+8x + 32) \\ 5 & 2500 & \gt 2000 \text{ no solution } \\ \ldots & \ldots & \gt 2000 \text{ no solution } \\ \hline \end{array}\)

All positive integers k  are 1,2,3,4

Suppose that the least common multiple of the first 25 positive integers is equal to

26A7114B4C0.

Find 100*A + 10*B + C.

\(\begin{array}{|rcll|} \hline && \text{lcm}(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25) \\ &=& 26{\color{red}7}7114{\color{red}4}4{\color{red}0}0 \\ && 26{\color{red}A}7114{\color{red}B}4{\color{red}C}0 \\\\ && A=7 \\ && B=4 \\ && C= 0\\ && 100*A + 10*B + C \\ &=& 100*7 + 10*4 + 0 \\ &=& 700 + 40 + 0 \\ &\mathbf{=}& \mathbf{740} \\ \hline \end{array} \)

I have no idea why the answer is what you say it is...How you get to it....thanx anyways...

\(\huge{ \begin{array}{|lrlrlrlrlrl|} \hline & a_{\color{red}1} &;& a_{\color{red}2}&;& a_{\color{red}3} &;& a_{\color{red}4} &;& \ldots &;& a_{\color{red}n} \\ \hline & \dfrac{1}{3}&;& \dfrac{8}{9}&;& 1&;& \dfrac{64}{81}&;& \ldots &;& \\\\ \Rightarrow & \dfrac{1^1}{3^1}&;& \dfrac{2^3}{3^2}&;& 1&;& \dfrac{4^3}{3^4}&;& \ldots &;& \\\\ \Rightarrow & \dfrac{{\color{red}1}^1}{3^{\color{red}1}}&;& \dfrac{{\color{red}2}^3}{3^{\color{red}2}}&; & \dfrac{{\color{red}3}^3}{3^{\color{red}3}}&;& \dfrac{{\color{red}4}^3}{3^{\color{red}4}}&;& \ldots &;&\dfrac{{\color{red}n}^3}{3^{\color{red}n}} \\ \hline \end{array} }\)

if we consider this sequence:

SAMKELLY

SAMKELLY

SAMKELLY....

The smiley is part of the sequence.

In what position will the 100th "L" be?

\(\begin{array}{|lrcll|} \hline &\boxed{ L_{2n} = 8+(n-1)\cdot 9 } \\\\ n=50: & L_{2\cdot 50}=L_{100} &=& 8+(50-1)\cdot 9 \\ & &=& 8+49\cdot 9 \\ & &=& 8+441\\ & &=& 449 \\ \hline \end{array}\)

The 100th "L" will be in position 449

A sequence of 2016 terms is formed as follows: The first two terms are equal to 3, from the 3rd term onwards, each consecutive term is formed by finding the sum of the previous two terms.

If each of the terms of the sequence 3,3,6,9....are now devided by 2,

and the remainders are added,

what will the sum of the first 2016 remainders be?

\(\begin{array}{|rcll|} \hline && \text{Let } \mathcal{F} \text{ are the Fibonacci numbers }\\\\ \mathcal{F}_1 &=& 1\\ \mathcal{F}_2 &=& 1\\ \mathcal{F}_3 &=& 2\\ \mathcal{F}_4 &=& 3\\ \mathcal{F}_5 &=& 4\\ \mathcal{F}_6 &=& 5\\ \mathcal{F}_7 &=& 13\\ \mathcal{F}_8 &=& 21\\ \mathcal{F}_9 &=& 34\\ \mathcal{F}_{10} &=& 55\\ \mathcal{F}_{11} &=& 89\\ \mathcal{F}_{12} &=& 144\\ \ldots \\ \hline \end{array} \)

The sequence:

\(\begin{array}{|rccr|l|} \hline a_1 &=& &=& 1\cdot 3 \\ a_2 &=& &=& 1\cdot 3 \\ a_3 &=& 1\cdot 3 + 1\cdot 3 &=& 2\cdot 3 \\ a_4 &=& 1\cdot 3 + 2\cdot 3 &=& 3\cdot 3 \\ a_5 &=& 2\cdot 3 + 3\cdot 3 &=& 5\cdot 3 \\ a_6 &=& 3\cdot 3 + 5\cdot 3 &=& 8\cdot 3 \\ a_7 &=& 5\cdot 3 + 8\cdot 3 &=& 13\cdot 3 \\ a_8 &=& 8\cdot 3 + 13\cdot 3 &=& 21\cdot 3 \\ a_9 &=& 13\cdot 3 + 21\cdot 3 &=& 34\cdot 3 \\ \ldots \\ a_n &=& && \mathcal{F}_{n} \cdot 3 \\ \hline \end{array} \)

\(\begin{array}{|r|rcl|c|} \hline \text{cycle} && && \text{remainder after divided by } 2 \\ \hline 1 & a_1 &=& 1\cdot 3 & 1 \\ 1 & a_2 &=& 1\cdot 3 & 1 \\ 1 & a_3 &=& 2\cdot 3 & 0 \\ \hline 2 & a_4 &=& 3\cdot 3 & 1 \\ 2 & a_5 &=& 5\cdot 3 & 1 \\ 2 & a_6 &=& 8\cdot 3 & 0 \\ \hline 3 & a_7 &=& 13\cdot 3 & 1 \\ 3 & a_8 &=& 21\cdot 3 & 1 \\ 3 & a_9 &=& 34\cdot 3 & 0 \\ \hline \ldots \\ \hline 672 & a_{2014} &=& \mathcal{F}_{2014} \cdot 3 & 1 \\ 672 & a_{2015} &=& \mathcal{F}_{2015} \cdot 3 & 1 \\ 672 & a_{2016} &=& \mathcal{F}_{2016} \cdot 3 & 0 \\ \hline &&&& \text{sum }=672*2 = 1344 \\ \hline \end{array} \)

The sum of the first 2016 remainders will be 1344

Hi good people!,
How do I go about finding the general term for the following sequence:
{1\over3};{8\over9};1;{64\over81}

\(\displaystyle {1\over3};{8\over9};1;{64\over81}\)

\(\huge{ \begin{array}{|rcll|} \hline a_1 &=& {1\over3} \\\\ a_2 &=& {8\over9} \\\\ a_3 &=& 1 \\\\ a_4 &=& {64\over81} \\\\ \ldots \\\\ a_n &=& \dfrac{n^3}{3^n} \\ \hline \end{array} }\)

The product of $3t^2+5t+a$ and $4t^2+bt-2$ is $12t^4+26t^3-8t^2-16t+6$. What is $a+b$?

\(\begin{array}{|rcll|} \hline && (3t^2+5t+a)(4t^2+bt-2) \\ &=& 12t^4+3bt^3-6t^2+20t^3+5bt^2-10t+4at^2+abt-2a \\ &=& 12t^4+(3b+20)t^3 +(-6+5b+4a)t^2 +(-10+ab)t -2a \\ \hline \text{compare}&=&12t^4+26t^3-8t^2-16t+6 \\\\ -2a &=& 6 \quad | \quad : (-2) \\ \mathbf{a} & \mathbf{=} & \mathbf{-3} \\\\ 3b+20 &=& 26 \\ 3b &=& 6 \\ \mathbf{b} & \mathbf{=} & \mathbf{2} \\\\ a+b &=& -3+2 \\ \mathbf{a+b} & \mathbf{=} & \mathbf{-1} \\ \hline \end{array} \)

Let  S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +............+ 500^2/(999.1001).

What is the sum of S?

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1^2}{1\cdot 3} + \dfrac{2^2}{3\cdot 5} + \dfrac{3^2}{5\cdot 7} + \dfrac{4^2}{7\cdot 9} + \ldots + \dfrac{500^2}{999\cdot 1001} \\\\ S &=& \sum \limits_{n=1}^{500} \dfrac{n^2}{(2n-1)(2n+1)} \\\\ && \boxed{\dfrac{1}{(2n-1)(2n+1)}=\dfrac12\cdot \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right)} \\\\ S &=& \sum \limits_{n=1}^{500} n^2\dfrac12 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\\\ S &=& \dfrac12 \sum \limits_{n=1}^{500} n^2 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\\\ S &=& \dfrac12 \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n-1} - \dfrac{n^2}{2n+1} \right) \\\\ 2S &=& \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n+1} \right) \\\\ 2S &=& \dfrac{1^2}{1\cdot2-1} + \sum \limits_{n=2}^{500} \left(\dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{2\cdot 500 + 1} \\\\ 2S &=& 1 + \sum \limits_{n=2}^{500} \left(\dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2}{2(n+1)-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2}{2n+1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2-n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{n^2+2n+1-n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{2n+1}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(1 \right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + 499 -\dfrac{500^2}{1001} \\\\ 2S &=& 500-\dfrac{500^2}{1001} \\\\ 2S &=& 500 \cdot \left(1-\dfrac{500}{1001} \right) \\\\ S &=& 250 \cdot \left(\dfrac{1001-500}{1001} \right) \\\\ S &=& \dfrac{250\cdot 501}{1001} \\\\ \mathbf{S} & \mathbf{=} & \mathbf{\dfrac{125250}{1001}} \\ \hline \end{array}\)

Find the value of $y$ such that $9^y = \dfrac{3^5\cdot 9^6}{27^3}$.

\(\begin{array}{|rcll|} \hline 9^y &=& \dfrac{3^5\cdot 9^6}{27^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{(3\cdot 9)^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{3^39^3} \\\\ 9^y &=& 3^{5-3}\cdot 9^{6-3} \\ 9^y &=& 3^{2}\cdot 9^{3} \\ 9^y &=& 9\cdot 9^{3} \\ 9^y &=& 9^1\cdot 9^{3} \\ 9^y &=& 9^{3+1} \\ 9^{{\color{red}y}} &=& 9^{{\color{red}4}} \\\\ \mathbf{y} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)