1. The graphs of y = x^3 - 3x + 2 and x + 4y = 4 intersect in the points , and . If and , compute the ordered pair (A,B).

1.
The graphs of $y = x^3 - 3x + 2 \text{ and } x + 4y = 4$ intersect in the points $(x_1,y_1), (x_2,y_2), \text{ and } (x_3,y_3)$.
If $x_1 + x_2 + x_3 = A \text{ and } y_1 + y_2 + y_3 = B$, compute the ordered pair $(A,B)$.

$\mathbf{\text{Vieta:}}$

$\begin{array}{rcll} x^3+ a_2x^2+a_1x+a_0 &=& 0 \qquad \mathbf{ a_2 =-(x_1+x_2+x_3)} \\ \end{array}$

$\begin{array}{|rcll|} \hline x + 4y &=& 4 \quad |\quad y = x^3 - 3x + 2 \\ x + 4(x^3 - 3x + 2) &=& 4 \\ 4x^3-11x+4 &=& 0 \quad |\quad : 4 \\ x^3-\dfrac{11}{4}x+1 &=& 0 \quad |\quad a_2 = 0~ !\\\\ \mathbf{x_1+x_2+x_3 } &=& \mathbf{0} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline x + 4y &=& 4 \\ x &=& 4-4y \\ \mathbf{x} &=& \mathbf{4(1-y)} \\\\ y &=& x^3 - 3x + 2 \quad |\quad \mathbf{x = 4(1-y)}\\ y &=& 64(1-y)^3 -3\cdot 4(1-y)+2 \\ \ldots \\ 64y^3-129y^2+181y-54 &=& 0 \quad |\quad : 64 \\ y^3-3y^2+\dfrac{181}{64}y-\dfrac{54}{64} &=& 0 \quad |\quad a_2 = -3~ !\\\\ -(y_1+y_2+y_3) &=& a_2 \\ -(y_1+y_2+y_3) &=& -3 \\ \mathbf{y_1+y_2+y_3 } &=& \mathbf{3} \\ \hline \end{array}$

$\mathbf{(A,B) = (0,3)}$

2.
Compute $\dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4}$

$\text{AP: $\quad a_n = a_1+(n-1)d,\qquad a_1=2,\ d=6 $}$

$\begin{array}{|rcll|} \hline a_n&=&2+(n-1)\cdot6\\\\ a_2 &=& 2+1\cdot 6 = 8 \\ a_3 &=& 2+2\cdot 6 \\ \ldots \\ a_{101} &=& 2+100\cdot 6 =602\\ \hline \end{array}$

$\text{GP: $\quad b_n = ar^{n-1},\qquad a=1,\ r=\dfrac{1}{4} $}$

$\begin{array}{|rcll|} \hline b_n &=& \left(\dfrac{1}{4}\right)^{n-1} \\\\ b_1 &=& \left(\dfrac{1}{4}\right)^{0} = 1 \\ b_2 &=& \dfrac{1}{4} \\ b_3 &=& \left(\dfrac{1}{4}\right)^{2} \\ \ldots \\ b_{101} &=& \left(\dfrac{1}{4}\right)^{100} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline s &=& \dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4} \\ \hline \end{array}$

$\begin{array}{|rclll|} \hline s &=& a_2b_{101}+& a_3b_{100}+a_4b_{99}+\ldots+ a_{101}b_{2} \\ \dfrac{s}{\frac{1}{4}} &=& & a_2b_{100}+ a_3b_{99}+\ldots+ a_{100}b_{2}+a_{101}b_1 \quad | \quad b_1 = 1,\ a_{n+1}-a_n = d \\ \hline s- \dfrac{s}{\frac{1}{4}} &=& a_2b_{101}+& d(b_2+b_3+\ldots + b_{100})-a_{101} \quad | \quad a_2 = 8,\ a_{101} = 602,\ d = 6 \\ -3s &=& 8b_{101}+& 6(\underbrace{b_2+b_3+\ldots + b_{100}}_{=S~ (GP)})-602 \\ -3s &=& 8b_{101}+& 6S-602 \\\\ &&&\begin{array}{|rclll|} \hline S &=& b_2+&b_3+\ldots + b_{100} \\ \dfrac{1}{4}S &=& & b_3+\ldots + b_{100}+b_{101} \\ \hline S - \dfrac{1}{4}S &=& b_2-& b_{101} \\ \dfrac{3}{4}S &=& b_2-& b_{101} \\ S &=& \dfrac{4}{3}b_2-& \dfrac{4}{3}b_{101} \\ \hline \end{array} \\\\ -3s &=& 8b_{101}+& 6\left(\dfrac{4}{3}b_2- \dfrac{4}{3}b_{101} \right)-602 \\ -3s &=& 8b_{101}+& 8b_2- 8b_{101} -602 \\ -3s &=& & 8b_2 -602 \quad | \quad b_2 = \dfrac{1}{4} \\ -3s &=& & 2 -602 \\ -3s &=& & -600 \quad | \quad : (-3) \\ \mathbf{ s} &=& &\mathbf{ 200 } \\ \hline \end{array}$

$\mathbf{\dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4} = 200}$

3.
A sequence $(a_n)$ is defined as follows: $a_1 = 1,\ a_2 = \dfrac{1}{2}$, and $a_n = \dfrac{1 - a_{n - 1}}{2a_{n - 2}}$ for all $n > 2$. Find $a_{120}$.

$a(1) = 1 \\ a(2) = 1/2 \\ a(3) = 1/4 \\ a(4) = 3/4 \\ a(5) = 1/2 \\ a(6) = 1/3 \\ a(7) = 2/3 \\ a(8) = 1/2 \\ a(9) = 3/8 \\ a(10) = 5/8 \\ a(11) = 1/2 \\ a(12) = 2/5 \\ a(13) = 3/5 \\ a(14) = 1/2 \\ a(15) = 5/12 \\ a(16) = 7/12 \\ a(17) = 1/2 \\ a(18) = 3/7 \\ a(19) = 4/7 \\ a(20) = 1/2 \\ a(21) = 7/16 \\ a(22) = 9/16 \\ a(23) = 1/2 \\ a(24) = 4/9 \\ a(25) = 5/9 \\ a(26) = 1/2 \\ a(27) = 9/20 \\ a(28) = 11/20 \\ a(29) = 1/2 \\ a(30) = 5/11 \\ a(31) = 6/11 \\ a(32) = 1/2 \\ a(33) = 11/24 \\ a(34) = 13/24 \\ a(35) = 1/2 \\ a(36) = 6/13 \\ a(37) = 7/13 \\ a(38) = 1/2 \\ a(39) = 13/28 \\ a(40) = 15/28 \\ a(41) = 1/2 \\ a(42) = 7/15 \\ a(43) = 8/15 \\ a(44) = 1/2 \\ a(45) = 15/32 \\ a(46) = 17/32 \\ a(47) = 1/2 \\ a(48) = 8/17 \\ a(49) = 9/17 \\ a(50) = 1/2 \\ a(51) = 17/36 \\ a(52) = 19/36 \\ a(53) = 1/2 \\ a(54) = 9/19 \\ a(55) = 10/19 \\ a(56) = 1/2 \\ a(57) = 19/40 \\ a(58) = 21/40 \\ a(59) = 1/2 \\ a(60) = 10/21 \\ a(61) = 11/21 \\ a(62) = 1/2 \\ a(63) = 21/44 \\ a(64) = 23/44 \\ a(65) = 1/2 \\ a(66) = 11/23 \\ a(67) = 12/23 \\ a(68) = 1/2 \\ a(69) = 23/48 \\ a(70) = 25/48 \\ a(71) = 1/2 \\ a(72) = 12/25 \\ a(73) = 13/25 \\ a(74) = 1/2 \\ a(75) = 25/52 \\ a(76) = 27/52 \\ a(77) = 1/2 \\ a(78) = 13/27 \\ a(79) = 14/27 \\ a(80) = 1/2 \\ a(81) = 27/56 \\ a(82) = 29/56 \\ a(83) = 1/2 \\ a(84) = 14/29 \\ a(85) = 15/29 \\ a(86) = 1/2 \\ a(87) = 29/60 \\ a(88) = 31/60 \\ a(89) = 1/2 \\ a(90) = 15/31 \\ a(91) = 16/31 \\ a(92) = 1/2 \\ a(93) = 31/64 \\ a(94) = 33/64 \\ a(95) = 1/2 \\ a(96) = 16/33 \\ a(97) = 17/33 \\ a(98) = 1/2 \\ a(99) = 33/68 \\ a(100) = 35/68 \\ a(101) = 1/2 \\ a(102) = 17/35 \\ a(103) = 18/35 \\ a(104) = 1/2 \\ a(105) = 35/72 \\ a(106) = 37/72 \\ a(107) = 1/2 \\ a(108) = 18/37 \\ a(109) = 19/37 \\ a(110) = 1/2 \\ a(111) = 37/76 \\ a(112) = 39/76 \\ a(113) = 1/2 \\ a(114) = 19/39 \\ a(115) = 20/39 \\ a(116) = 1/2 \\ a(117) = 39/80 \\ a(118) = 41/80 \\ a(119) = 1/2 \\ a(120) = 20/41$

2 - ∑[(6*n + 2) / 4^(100 - n +1), n , 1, 100] = 200

3 - This sequence has this recurrence relation:
a(n + 3) =((n + 2) a(n))/(n + 6) + a(n + 1)/(-n - 6) + a(n + 2)/(-n - 6) + 3/(n + 6): Where:
a(1) = 1
a(2) = 1/2
a(3) = 1/4
a(4) = 3/4
.
.
.
.
a(120) = 1/123 (119 a(117) - a(118) - a(119) + 3)
a(120) = 20 / 41

3.
A sequence $(a_n)$ is defined as follows: $a_1 = 1,\ a_2 = \dfrac{1}{2}$, and $a_n = \dfrac{1 - a_{n - 1}}{2a_{n - 2}}$ for all $n > 2$. Find $a_{120}$.

$a_n =\begin{cases} \dfrac{1}{2}, & \text{if }n\equiv 2 \pmod{3} \\\\ \dfrac{n}{2n+6}, & \text{if }n\equiv 0 \pmod{3} \\\\ \dfrac{n+5}{2n+4}, & \text{if }n\equiv 1 \pmod{3} \\ \end{cases}$

$\begin{array}{|rcll|} \hline n &=& 120 \quad \text{ and }\quad 120 &\equiv& 0 \pmod{3} \\ \\ a_{120} &=& \dfrac{n}{2n+6} \\\\ &=& \dfrac{120}{2\cdot 120+6} \\\\ &=& \dfrac{120}{246} \\\\ \mathbf{a_{120}} &=& \mathbf{\dfrac{20}{41}} \\ \hline \end{array}$