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.

escumantsu

Step 1
Given that: $\left\{{a}_{n}\right\}$ be a sequance of defined recursively
${a}_{n}=3{a}_{n-1}-2{a}_{n-2}$ for $n\ge 2$
To show: ${a}_{n}=3\cdot {2}^{n}-3$ for $n\ge 2$
we will use strong mathematical induction
for

So, for it is true
Let us assume it is true for all $k\le n$
Such that
Now for $k=n+1$
So, ${a}_{n+1}=3{a}_{n}-2{a}_{n-1}$
$=3\left(3\cdot {2}^{n}-3\right)-2\left(3\cdot {2}^{n-1}-3\right)$
$\left(\because$ by induction
${a}_{n+1}={3}^{2}\cdot {2}^{n}-9-6\cdot {2}^{n-1}+6$
$=3\cdot {2}^{n-1}\left(3\cdot 2-2\right)-3$
$=3\cdot {2}^{n-1}\cdot 4-3$
$=3\cdot {2}^{n-1}\cdot {2}^{2}-3$
${a}_{n+1}=3\cdot {2}^{n+1}-3$

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