Show that the sequence an is an solution of the recurrence relation an=−3an−1+4an−2 if an=2(−4)n+3.

Question

Show that the sequence  an is an solution of the recurrence relation an=−3an−1+4an−2 if an=2(−4)n+3.

izboknil3 · Accepted Answer

Given: an=2(−4)n+3 n=0,1,2,... To prove: an=−3an−1+4an−2,n≥2. Proof Replace n in an=2(−4)n+3 by n−1: an−1=2(−4)n−1+3 Replace n in an=2(−4)n+3 by n−2: an−2=2(−4)n−2+3 We will start from the expression −3an−1+4an−2 and prove that this term has to be equal to an(when n≥2). Let us use the two previous expressions derived for an−1 and an−2 −3an−1+4an−2=−3(2(−4)n−1+3)+4(2(−4)n−2+3) Use distributive property: =−6(−4)n−1−9+8(−4)n−2+12 Group the powers of -4: =(−6(−4)n−1+8(−4)n−2)−9+12 =(−6(−4)⋅(−4)n−2+8(−4)n−2)+3 Let us factor out (−4)n−2: =(−4)n−2(−6⋅(−4)+8)+3 =(−4)n−2(24+8)+3 =(−4)n−2⋅32+3 =(−4)n−2⋅16⋅2+3 =(−4)n−2⋅(−4)2⋅2+3 =(−4)n−2⋅2+3

Show that the sequence a_{n} is an solution of the recurrence relation PS

Answered question

Answer & Explanation

New Questions in Algebra I