Recurrence relation nonhomogeneous recursion: An + B = 7 (A (n-1) + B) - 10 ( A

Question

Recurrence relation nonhomogeneous recursion: An + B = 7 (A (n-1) + B) - 10 ( A (n-2) + B + n = 7 An-7A + 7B - 10An + 20 A - 10B +N or (4A -1)N -13A + 4B=0.. This is the ANSWER A=1/4 and B =13/16. Can someone explain how 7An-7A+7B -10An... can turn into= (4A-1)n -13A+4B=0..(simplification? I cannot understand this part of the problem's answer because the instructor skipped some steps to end with an answer A=1/4 and B 13/16.. Can you solve this without skipping parts of your solution? or can you solve the An+B = 7 (A (n-1) +B ... and arrive with the answer for A and B given above????

Explanation / Answer

An + B = 7 (A (n-1) + B) - 10 ( A (n-2) + B) + n

           = 7 An-7A + 7B - 10An + 20 A - 10B +n

what he has done was taken lhs terms to rhs and formed the final equation as

7 An-7A + 7B - 10An + 20 A - 10B +n -An - B = 0

=> -4An + 13A -4B + n =0

=> 4An + n -13A +4B =0

=> (4A - 1)n - 13A + 4B = 0

=> 4A-1 = 0 and -13A + 4B = 0

=> A = 1/4 and B = 13A/4 => B= 13/16

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