Ava-May Nelson

2021-08-16

We have a recursively defined sequence $a}_{n$ .

${a}_{0}=0,{a}_{1}=3$ , and $a}_{n}=3{a}_{n-1}-2{a}_{n-2$ for $n\ge 2$

We would like to prove that f or all$n\ge 0,{a}_{n}=3\cdot {2}^{n}-3$ .

Prove this using the stronger mathematical induction.

We would like to prove that f or all

Prove this using the stronger mathematical induction.

escumantsu

Skilled2021-08-17Added 98 answers

Step 1

Given that:$\left\{{a}_{n}\right\}$ be a sequance of defined recursively ${a}_{0}=0,\text{}{a}_{1}=3$

$a}_{n}=3{a}_{n-1}-2{a}_{n-2$ for $n\ge 2$

To show:${a}_{n}=3\cdot {2}^{n}-3$ for $n\ge 2$

we will use strong mathematical induction

for$n=0\text{}{a}_{0}=3\cdot {2}^{0}-3=0$

$n=1\text{}{a}_{1}=3\cdot {2}^{1}-3=3$

So, for$n=0,\text{}n=1$ it is true

Let us assume it is true for all$k\le n$

Such that${a}_{k}=3\cdot {2}^{k}-3\text{}\mathrm{\forall}k\le n$

Now for$k=n+1$

So,$a}_{n+1}=3{a}_{n}-2{a}_{n-1$

$=3(3\cdot {2}^{n}-3)-2(3\cdot {2}^{n-1}-3)$

$(\because$ by induction ${a}_{n}=3\cdot {2}^{n}-3,\text{}{a}_{n-1}=3\cdot {2}^{n-1}-3)$

${a}_{n+1}={3}^{2}\cdot {2}^{n}-9-6\cdot {2}^{n-1}+6$

$=3\cdot {2}^{n-1}(3\cdot 2-2)-3$

$=3\cdot {2}^{n-1}\cdot 4-3$

$=3\cdot {2}^{n-1}\cdot {2}^{2}-3$

${a}_{n+1}=3\cdot {2}^{n+1}-3$

Given that:

To show:

we will use strong mathematical induction

for

So, for

Let us assume it is true for all

Such that

Now for

So,