heureka

The function f(n) = 3f(n-2) - 2f(n-1), where f(2) = 3 and f(1) = -1. What is the value of f(5)?

Formula:

\(\begin{array}{|rcll|} \hline f_n &=& -2f_{n-1}+3f_{n-2} \quad & | \quad f_2 = 3 \qquad f_1 = -1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f_3 &=& -2f_{2}+3f_{1} \quad & | \quad f_2 = 3 \qquad f_1 = -1 \\ &=& -2(3)+3(-1) \\ &=& -6-3 \\ \mathbf{f_3} & \mathbf{=} & \mathbf{-9} \\\\ \hline f_4 &=& -2f_{3}+3f_{2} \quad & | \quad f_3 = -9 \qquad f_2 = 3 \\ &=& -2(-9)+3(3) \\ &=& 18+9 \\ \mathbf{f_4} & \mathbf{=} & \mathbf{27} \\\\ \hline f_5 &=& -2f_{4}+3f_{3} \quad & | \quad f_4 = 27 \qquad f_3 = -9 \\ &=& -2(27)+3(-9) \\ &=& -54 -27 \\ \mathbf{f_5} & \mathbf{=} & \mathbf{-81} \\ \hline \end{array}\)

\(\mathbf{f(5) = -81}\)

Formula:

\(\begin{array}{|lrcll|} \hline & f_n &=& -2f_{n-1}+3f_{n-2} \\ \text{ or } \\ & f_n &=& -(-3)^{n-1} \\ & f_5 &=& -(-3)^{5-1} \\ & f_5 &=& -(-3)^{4} \\ & \mathbf{f_5} & \mathbf{=} & \mathbf{-81} \\ \hline \end{array}\)

Formula edited, thank you Alan!

\(\mathbf{f(5) = -81}\)

Find all complex numbers  such that  z^2=2i

Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer

\(\begin{array}{|lrcll|} \hline & (a+bi)^2 &=& a^2 + 2abi+b^2i^2 \quad | \quad i^2 = -1 \\ & (a+bi)^2 &=& a^2 + 2abi -b^2 \\ & (a+bi)^2 &=& \underbrace{a^2-b^2}_{=0} + \underbrace{2abi}_{=2i} \quad | \quad \text{compare with } 2i \\\\ 1. & a^2-b^2 &=& 0 \\ & a^2 &=& b^2 \\ & \mathbf{a} & \mathbf{=}& \mathbf{b} \\\\ 2. & 2abi &=& 2i \quad | \quad : 2i \\ &\mathbf{ ab } & \mathbf{=}& \mathbf{1 } \quad | \quad a=b \\ &a^2 & = & 1 \\ &a & = & \pm 1 \\ \\ &b &=& a \\ & b &=& \pm 1 \\\\ & && & \mathbf{ab = 1\ ?} \\ & a &=& 1 \quad \text{ and } \quad b = 1 & ab = 1\cdot 1 = 1\ \checkmark \\ \text{or}& a &=& -1 \quad \text{ and } \quad b = -1 & ab = (-1)\cdot (-1) = 1\ \checkmark \\ \text{else}& a &=& -1 \quad \text{ and } \quad b = 1 & ab = (-1)\cdot 1 = -1\ \text{ no solution} \\ & a &=& 1 \quad \text{ and } \quad b = -1 & ab = 1\cdot (-1) = -1\ \text{ no solution} \\ \hline \end{array}\)

The answer is \(\mathbf{"1+i, -1-i"}\)

How many positive multiples of 7 that are less than 1000 end with the digit 3?

\(\begin{array}{|r|r|r|} \hline & n & 7n \\ \hline 1. &9& 63\\ 2. &19& 133\\ 3. &29& 203\\ 4.&39& 273\\ 5.&49& 343\\ 6.&59& 413\\ 7.&69& 483\\ 8.&79& 553\\ 9.&89& 623\\ 10.&99& 693\\ 11.&109& 763\\ 12.&119& 833\\ 13.&129& 903\\ 14.&139& 973\\ \hline \end{array}\)

14 positive multiples of 7 that are less than 1000 end with the digit 3.

Find all solutions of the equation(answers)

|x^2 - 14x + 29| = 4

\(\begin{array}{|lrclcrcl|} \hline & |x^2 - 14x + 29| &=& 4 \\ \Rightarrow & |(x-7)^2-20| &=& 4 \quad & | \quad \text{substitute} \quad u = (x-7)^2 \\\\ & |u-20| &=& 4 \\ & u_1-20 &=& 4 &\text{ or }& u_2-20 &=& -4 \\ & u_1 &=& 24 && u_2 &=& 16 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline u &=& (x-7)^2 \\ \pm\sqrt{u} &=& x-7 \\ 7 \pm\sqrt{u} &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{7 \pm\sqrt{u}} \\ \hline \end{array}\)

Solutions:

\(\begin{array}{|rclcl|} \hline x_1 &=& 7+\sqrt{u_1} = 7 + \sqrt{24} &=& 7+2\sqrt{6} \\ x_2 &=& 7-\sqrt{u_1} = 7 - \sqrt{24} &=& 7-2\sqrt{6} \\ x_3 &=& 7+\sqrt{u_2} = 7 + \sqrt{16} &=& 11 \\ x_4 &=& 7-\sqrt{u_2} = 7 - \sqrt{16} &=& 3 \\ \hline \end{array}\)

Proof:

\(\begin{array}{|rcll|} \hline |(7+2\sqrt{6})^2 - 14\cdot (7+2\sqrt{6}) + 29| &=& 4 \\ |11.8989794856^2 - 14\cdot 11.8989794856 + 29| &=& 4 \\ |-25 + 29| &=& 4 \\ |4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |(7-2\sqrt{6})^2 - 14\cdot (7-2\sqrt{6}) + 29| &=& 4 \\ |2.10102051443^2 - 14\cdot 2.10102051443 + 29| &=& 4 \\ |-25 + 29| &=& 4 \\ |4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |11^2 - 14\cdot 11 + 29| &=& 4 \\ |121 - 154 + 29| &=& 4 \\ |-33 + 29| &=& 4 \\ |-4| &=& 4 \\ 4 &=& 4 \ \checkmark \\\\ \hline |3^2 - 14\cdot 3 + 29| &=& 4 \\ |9 - 42 + 29| &=& 4 \\ |-33 + 29| &=& 4 \\ |-4| &=& 4 \\ 4 &=& 4 \ \checkmark \\ \hline \end{array}\)

3)
(1/4)x^2 + 12=0

\(\begin{array}{|rcll|} \hline (1/4)x^2 + 12 &=& 0 \quad | \quad \cdot 4 \\ x^2 + 48 &=& 0 \quad | \quad -48 \\ x^2 &=& -48 \\ x^2 &=& (-1)\cdot 16 \cdot 3 \\ x &=& \pm \sqrt{(-1)\cdot 16 \cdot 3} \\ x &=&\pm\sqrt{16}\sqrt{3}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm 4\sqrt{3}\cdot i \\ \hline \end{array}\)

2)
6x^2 = -126

\(\begin{array}{|rcll|} \hline 6x^2 &=& -126 \quad | \quad : 6 \\ x^2 &=& -21 \\ x^2 &=& (-1)\cdot 21 \\ x &=& \pm\sqrt{(-1)\cdot 21} \\ x &=&\pm \sqrt{21}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm \sqrt{21}\cdot i \\ \hline \end{array}\)

1)

3x^2 + 81=0

\(\begin{array}{|rcll|} \hline 3x^2 + 81 &=& 0 \quad | \quad : 3 \\ x^2 + 27 &=& 0 \quad | \quad -27 \\ x^2 &=& -27 \\ x^2 &=& (-1)\cdot 9 \cdot 3 \\ x &=& \pm\sqrt{(-1)\cdot 9 \cdot 3} \\ x &=&\pm\sqrt{9}\sqrt{3}\sqrt{-1} \quad &| \quad \sqrt{-1} = i \\ x &=&\pm 3\sqrt{3}\cdot i \\ \hline \end{array}\)

John counts up from 1 to 13, and then immediately counts down again to 1, and then back up to 13, and so on, alternately counting up and down: What is the 5000th integer in his list?

Variation 1:

\([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,][13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1] \ldots\)

\(\text{position $= n \pmod{26} \qquad$ position $0 =$ position $26$} \\ \text{$5000th \pmod {26} =$ position $8$}\)

Variation 2:

\([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,][ 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2 ] \ldots\)

\(\text{position $= n \pmod{24} \qquad$ position $0 =$ position $24$} \\ \text{$5000th \pmod {24} =$ position $8$}\)

Let {a_n} is a 3rd order arithmetic sequence. The first couple of terms are: 4, 16, 42, 88.
Please determine the general formula of {a_n}

\(\begin{array}{lrrrrrrrrrr} & {\color{red}d_0 = 4} && 16 && 42 && 88 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = 12} && 26 && 46 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 14} && 20 && \cdots \\ \text{3. Difference } &&&& {\color{red}d_3 = 6} && \cdots \\ \end{array}\)

\(\begin{array}{rcl} a_n &=& \binom{n-1}{0}\cdot {\color{red}d_0 } + \binom{n-1}{1}\cdot {\color{red}d_1 } + \binom{n-1}{2}\cdot {\color{red}d_2 } + \binom{n-1}{3}\cdot {\color{red}d_3 } \\\\ a_n &=& \binom{n-1}{0}\cdot {\color{red} 4 } + \binom{n-1}{1}\cdot {\color{red}12} + \binom{n-1}{2}\cdot {\color{red}14} + \binom{n-1}{3}\cdot {\color{red} 6} \\\\ \hline \binom{n-1}{0} &=& 1 \\ \binom{n-1}{1} &=& n-1 \\ \binom{n-1}{2} &=& \left( \frac{n-1}{2} \right) \cdot \left( \frac{n-2}{1} \right) \\ \binom{n-1}{3} &=& \left( \frac{n-1}{3} \right) \cdot \left(\frac{n-2}{2} \right) \cdot \left( \frac{n-3}{1} \right) \\ \hline \\ a_n &=& {\color{red}4 } + (n-1)\cdot {\color{red}12} + \left( \frac{n-1}{2} \right) \cdot \left( \frac{n-2}{1} \right)\cdot {\color{red}14} + \left( \frac{n-1}{3} \right) \cdot \left(\frac{n-2}{2} \right) \cdot \left( \frac{n-3}{1} \right)\cdot {\color{red}6} \\\\ &=& 4 + (n-1)\cdot 12 +(n-1)(n-2) \cdot 7 + (n-1)(n-2)(n-3) \\\\ &=& 4 + 12n-12 + (7n-7)(n-2) + (n-1)(n-2)(n-3) \\\\ &=& -8 + 12n + 7n^2-21n +14 + (n-1)(n-2)(n-3) \\\\ &=& 6 - 9n + 7n^2 + (n-1)(n-2)(n-3) \\\\ &=& 6 - 9n + 7n^2 + (n-1)(n^2-5n+6) \\\\ &=& 6 - 9n + 7n^2 + n^3-5n^2+6n-n^2+5n-6 \\\\ \mathbf{a_n} &=& \mathbf{2n + n^2 + n^3} \\ \end{array} \)

If AAA_4 can be expressed as 33_b, where A is a digit in base 4 and b is a base greater than 5,

what is the smallest possible sum A+b?

\(\text{$AAA_4 = 33_b$ } \\ \text{Let $A = \{0,1,2,3\}$ }\)

\(\begin{array}{|rcll|} \hline A\cdot 4^2+A\cdot 4^1 + A\cdot 4^0 &=& 3b^1 + 3b^0 \\ A\cdot 16+A\cdot 4 + A &=& 3b + 3 \\ 21A &=& 3\cdot (b + 1) \quad & | \quad : 3 \\ 7A &=& b + 1 \\ \mathbf{b} &\mathbf{=}& \mathbf{7A -1} \quad & | \quad A = \{0,1,2,3\} \\ \hline \end{array} \)

\(\begin{array}{|r|r|c|r|} \hline A & \mathbf{b=7A -1} & b \gt 5 & A+b \\ \hline 0 & -1 & & -1 \\ \hline 1 & 6 & \checkmark & \color{red}7 \\ \hline 2 & 13 & \checkmark & 15 \\ \hline 3 & 20 & \checkmark & 23 \\ \hline \end{array}\)

\(\text{$b \gt 5$ and the smallest possible sum $\mathbf{A+b}$ is $1+6 \mathbf{=7} $ }\)

\(111_4 = 33_6\)