Recurrence Relation where n+1th Term is A by nth term + B to the n

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Theorem

Let $\sequence {a_n}$ be the sequence defined by the recurrence relation:

$a_n = \begin {cases} 0 & : n = 0 \\ A a_{n - 1} + B^{n - 1} & : n > 0 \end {cases}$

for numbers $A$ and $B$.


Then the closed form for $\sequence {a_n}$ is given by:

$a_n = \begin {cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end {cases}$


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$a_n = \begin {cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end {cases}$


$\map P 0$ is the case:

\(\ds \dfrac {A^0 - B^0} {A - B}\) \(=\) \(\ds \dfrac {1 - 1} {A - B}\)
\(\ds \) \(=\) \(\ds \dfrac 0 {A - B}\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds a_0\)


When $A = B$:

\(\ds 0 A^{-1}\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds a_0\)


Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds \dfrac {A^1 - B^1} {A - B}\) \(=\) \(\ds \dfrac {A - B} {A - B}\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds 0 \times A + 1\)
\(\ds \) \(=\) \(\ds A a_0 + B^0\)
\(\ds \) \(=\) \(\ds a_1\)


When $A = B$:

\(\ds 1 \times A^0\) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds A \times 0 + 1\)
\(\ds \) \(=\) \(\ds A \times a_0 + A^0\)


Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


This is the induction hypothesis:

$a_k = \begin {cases} \dfrac {A^k - B^k} {A - B} & : A \ne B \\ k A^{k - 1} & : A = B \end {cases}$


from which it is to be shown that:

$a_{k + 1} = \begin {cases} \dfrac {A^{k + 1} - B^{k + 1} } {A - B} & : A \ne B \\ \paren {k + 1} A^k & : A = B \end {cases}$


Induction Step

This is the induction step:


First let $A \ne B$.

\(\ds a_{k + 1}\) \(=\) \(\ds A a_k + B^k\) by definition
\(\ds \) \(=\) \(\ds A \dfrac {A^k - B^k} {A - B} + B^k\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \dfrac {A \paren {A^k - B^k} + \paren {A - B} B^k} {A - B}\)
\(\ds \) \(=\) \(\ds \dfrac {A^{k + 1} - A B^k + A B^k - B^{k + 1} } {A - B}\)
\(\ds \) \(=\) \(\ds \dfrac {A^{k + 1} - B^{k + 1} } {A - B}\)


When $A = B$ we have for $k > 0$:

\(\ds a_{k + 1}\) \(=\) \(\ds A a_k + A^k\)
\(\ds \) \(=\) \(\ds A \paren {k A^{k - 1} } + A^k\)
\(\ds \) \(=\) \(\ds k A^k + A^k\)
\(\ds \) \(=\) \(\ds \paren {k + 1} A^k\)

So $\map P k \implies \map P {k + 1}$ and by the Principle of Mathematical Induction:

$a_n = \begin {cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end {cases}$

$\blacksquare$


Sources

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