We have a recursively defined sequence an. a0=0,a1=3, and an=3an−1−2an−2 for n≥2 We would like to prove that f or all n≥0,an=3⋅2n−3. Prove this using the stronger mathematical induction.

Name: We have a recursively defined sequence an. a0=0,a1=3, and an=3an−1−2an−2 for n≥2 We would like to prove that f or all n≥0,an=3⋅2n−3. Prove this using the stronger mathematical induction.
Uploaded: 2022-02-23T17:22:57+00:00
Description: We have a recursively defined sequence an. a0=0,a1=3, and an=3an−1−2an−2 for n≥2 We would like to prove that f or all n≥0,an=3⋅2n−3. Prove this using the stronger mathematical induction.

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We have a recursively defined sequence an.  a0=0,a1=3, and an=3an−1−2an−2 for n≥2  We would like to prove that f or all n≥0,an=3⋅2n−3.  Prove this using the stronger mathematical induction.

escumantsu · Accepted Answer

Step 1  Given that: {an} be a sequance of defined recursively a0=0, a1=3  an=3an−1−2an−2 for n≥2  To show: an=3⋅2n−3 for n≥2  we will use strong mathematical induction  for n=0 a0=3⋅20−3=0  n=1 a1=3⋅21−3=3  So, for n=0, n=1 it is true  Let us assume it is true for all k≤n  Such that ak=3⋅2k−3 ∀k≤n  Now for k=n+1  So, an+1=3an−2an−1  =3(3⋅2n−3)−2(3⋅2n−1−3)  (∵ by induction an=3⋅2n−3, an−1=3⋅2n−1−3)  an+1=32⋅2n−9−6⋅2n−1+6  =3⋅2n−1(3⋅2−2)−3  =3⋅2n−1⋅4−3  =3⋅2n−1⋅22−3  an+1=3⋅2n+1−3

We have a recursively defined sequence an. a0=0,a1=3, and an=3an−1−2an−2 for n≥2 We would like...

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