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• 2009 AIME II Problems/Problem 14
  The new recurrence then becomes <math>b_0=0</math> and <math>b_{n+1} = \frac45 b_n + \frac 35\
  4 KB (680 words) - 13:49, 23 December 2023
• Characteristic polynomial
  ...h> be a sequence of real numbers. Consider a monic [[homogenous]] [[linear recurrence]] of the form ...dots, c_k</math> are real constants. The characteristic polynomial of this recurrence is defined as the polynomial
  19 KB (3,412 words) - 14:57, 21 September 2022
• 2010 USAMO Problems/Problem 3
  recurrence relations:
  11 KB (1,889 words) - 13:45, 4 July 2013
• Mock AIME 1 Pre 2005 Problems/Problem 13
  A sequence <math> \{R_{n}\}_{n\ge 0} </math> obeys the recurrence <math> 7R_{n}= 64-2R_{n-1}+9R_{n-2} </math> for any integers <math>n\ge 2<
  1 KB (178 words) - 14:22, 26 March 2011
• Linear recurrence
  ...ence <math>\{a_{0},a_{1},a_{2},\ldots\}</math> is said to obey a '''linear recurrence of order <math>k</math>''' if there exist constants <math>c_{0},c_{1},\ldot
  315 bytes (56 words) - 23:32, 2 June 2011
• 1990 AHSME Problems/Problem 30
  ...th> and <math>R_1=3</math>, all terms are integers and we can continue the recurrence <math>\rm{mod}\ 10</math> to get the repeating sequence <math>1,3,7,9,7,3</
  1,010 bytes (168 words) - 01:15, 5 February 2016
• 2013 USAMO Problems/Problem 2
  For the sake of convenience in determining recurrence relations, we define another type of board with two <math>1\times n</math>
  14 KB (2,076 words) - 20:29, 10 July 2023
• 1972 IMO Problems/Problem 3
  ...ntegral for all <math>m,n \geq 0</math>. To start, we would like to find a recurrence relation for <math>f(m,n)</math>. First, let's look at <math>f(m,n-1)</math Therefore, we have found the recurrence relation <cmath>f(m,n)=4f(m,n-1)-f(m+1,n-1).</cmath>
  3 KB (566 words) - 00:26, 7 December 2022
• 1964 AHSME Problems/Problem 31
  ...ula). And <math>f(n)</math> is the solution (i.e. <math>a_n</math>) to the recurrence relation with constants <math>a = \frac{5+3\sqrt{5}}{10}</math> and <math>b
  3 KB (419 words) - 23:15, 29 June 2023
• 1992 AHSME Problems/Problem 29
  ...y for the very hand-wavy and non-rigorous answer! At least mine has a cool recurrence relation! :D
  3 KB (472 words) - 19:29, 20 June 2021
• 2015 AMC 12A Problems/Problem 22
  One method of approach is to find a recurrence for <math>S(n)</math>. ...h>B</math> of length <math>n-1, n-2, \text{or}\ n-3</math>. So we have the recurrence: <cmath>A(n) = B(n-1) + B(n-2) + B(n-3) = A(n-1) + A(n-2) + A(n-3)</cmath>
  5 KB (925 words) - 13:08, 5 May 2024
• 2015 AIME II Problems/Problem 10
  The simple recurrence can be found.
  2 KB (245 words) - 11:43, 20 December 2021
• 2015 AIME II Problems/Problem 14
  ...= 810/z^4</math>. We desire <math>2P_3 + z^3</math>. Now building up this recurrence;
  10 KB (1,751 words) - 22:21, 26 November 2023
• 2016 AMC 12B Problems/Problem 25
  ...for all <math>n\geq 2</math>. The characteristic polynomial of this linear recurrence is <math>x^2-x-2=0</math>, which has roots <math>2</math> and <math>-1</mat
  4 KB (735 words) - 19:10, 11 January 2024
• 2016 AIME II Problems/Problem 12
  ...icity, denote segment <math>1</math>'s color as red), we get the following recurrence relation:
  20 KB (3,328 words) - 21:13, 5 April 2024
• 2017 AIME I Problems/Problem 9
  Taking the recurrence mod <math>99</math>, we have <cmath>a_n=a_{n-1}+n</cmath> for <math>a_{10}=
  10 KB (1,615 words) - 22:37, 16 December 2023
• 2018 AIME I Problems/Problem 11
  Using our proof, it is easy to see that <math>f(n)</math> satisfies the recurrence <math>f(1) = 2</math> and <math>f(n+1) = f(n) + 2</math>. Since this implie
  10 KB (1,448 words) - 06:30, 21 April 2024
• 2018 USAJMO Problems/Problem 1
  ...math>a_2 = 32</math>. We claim that for <math>n \geq 3</math> we have the recurrence <math>a_n = 8a_{n-1} + 9a_{n-2}</math>. This gives us a linear three-term recurrence. It is well-known that its solution is of the form <math>c_1r_1^n + c_2r_2
  10 KB (1,799 words) - 11:37, 28 February 2023
• 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 10
  Finally the implicit recurrence in (d) is obtained by running Newton backwards
  3 KB (419 words) - 04:57, 19 January 2019
• 2019 AMC 12A Problems/Problem 25
  ...ve for <math>p</math>, <math>q</math>, and <math>r</math> by iterating the recurrence to obtain <math>x_1=180^\circ-2x_0</math>, <math>x_2=4x_0-180^\circ</math>, ...ath> assuming our inductive hypothesis holds for <math>n</math>. Using the recurrence relation, we have <cmath>\begin{align*}
  5 KB (933 words) - 22:23, 2 January 2024

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