Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 4"

Latest revision as of 20:08, 11 July 2021

Problem

Let $(x_n)_{n\geq 0}$ and $(y_n)_{n\geq 0}$ be sequences of real numbers such that $x_0 = 3$ , $y_0 = 1$ , and, for all positive integers $n$ ,

$x_{n+1}+y_{n+1} = 2x_n + 2y_n,$ $x_{n+1}-y_{n+1}=3x_n-3y_n.$ Find $x_5$ .

Solution

We notice that $x_5 + y_5 = 2(x_4 + y_4)$ $= 2(2(x_3 + y_3))$ $= 2(2(2(x_2 + y_2)))$ $= 2(2(2(2(x_1 + y_1))))$ $= 2(2(2(2(2(x_0 + y_0))))).$ Since we are given that $x_0 = 3$ and $y_0 = 1$ , we can plug these values in to get that $x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128 \qquad (1).$

Similarly, we conclude that $x_5 - y_5 = 3(3(3(3(3(x_0 - y_0)))))= 3(3(3(3(3(3 - 1))))) = 3(3(3(3(3(2))))) = 486 \qquad (2).$

Adding $(1)$ and $(2)$ gives us $2 \cdot x_5 = 614.$ Dividing both sides by $2$ yields $x_5 = \boxed{307}.$

~mahaler

Solution 2

Add both equations to get $2(x_{n+1})=5x_n-y_n$ , and subtract both equations to get $2(y_{n+1})=5y_n-x_n$ , so now we bash: $x_1=7$ and $y_1=1$ . $x_2=17$ and $y_2=-1$ . $x_3=43$ and $y_3=-11$ . $x_4=113$ and $y_4=-49$ , $x_5=\frac{614}{2}=\boxed{307}$

~Geometry285

@@ Line 15: / Line 15: @@
 ~mahaler
+==Solution 2==
+Add both equations to get <math>2(x_{n+1})=5x_n-y_n</math>, and subtract both equations to get <math>2(y_{n+1})=5y_n-x_n</math>, so now we bash: <math>x_1=7</math> and <math>y_1=1</math>. <math>x_2=17</math> and <math>y_2=-1</math>. <math>x_3=43</math> and <math>y_3=-11</math>. <math>x_4=113</math> and <math>y_4=-49</math>, <math>x_5=\frac{614}{2}=\boxed{307}</math>
+~Geometry285
 ==See also==

Page

Toolbox

Search

Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 4"

Latest revision as of 20:08, 11 July 2021

Contents

Problem

Solution

Solution 2

See also