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How to solve the recurrence relation $a_n - 2 a_{n-1} = 3 \times 2^n, a_0 = 1$. By looking at the terms of the relation, it can be seen that it is linear in nature but it is not homogeneous. How to solve such a recurrence relation?

ephemeral
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6 Answers6

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Hint: let $\,a_n=2^n b_n\,$, then $\,2^n b_n - 2 \cdot 2^{n-1} b_{n-1} = 3 \cdot 2^{n} \iff b_n - b_{n-1}=3\,$, so $b_n$ is an arithmetic progression, with common difference $\,3\,$ and $\,b_0=a_0 / 2^0 = 1\,$.

dxiv
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  • What's the general principle for when such a substitution will work? – G Tony Jacobs Aug 28 '17 at 15:05
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    @GTonyJacobs The homogeneous recurrence has the general solution $,a_n=c 2^n,$. One possible heuristic is to replace the constant $,c,$ with a function on $,n,$ i.e. $,a_n=b_n 2^n,$ and try to find $,b_n,$ so that the full, non-homogeneous recurrence is satisfied. It is not a "general principle" and it doesn't always work out smoothly, but it did in this case. – dxiv Aug 28 '17 at 16:14
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    @GTonyJacobs P.S. More generally, any recurrence in the form $Aa_n+Ba_{n-1}=CD^n$ reduces to a linear one with constant coefficients via the substitution $a_n=D^nb_n,$, which gives $Ab_n+\frac{B}{D}b_{n-1}=C,$. – dxiv Aug 28 '17 at 17:21
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Considering the corresponding homogenous recurrence, a general solution for it is $a_n=A\cdot 2^n$. Since the inhomogeneous term resembles the homogeneous solution, we try adding a term of the form $B\cdot n\cdot 2^n$.

Taking $a_n=A\cdot 2^n + B\cdot n\cdot 2^n$, together with $a_0=1, a_1=8$, we obtain

$$A=1\\ 2A+2B=8$$

So our general term is $a_n=2^n+3n2^n$, or $a_n=(3n+1)2^n$.

G Tony Jacobs
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Hint

The first thing I would have is to set $a_n=2^n b_n$ which makes $$a_n - 2 a_{n-1} = 3 \times 2^n\implies b_n-b_{n-1}=3$$ which looks to be simple.

Claude Leibovici
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Here is a more general approach that will allow you to solve a variety of such problems. Consider that

$$ f_n-2f_{n-1}=3\cdot2^n\\ f_{n-1}-2f_{n-2}=\frac{3\cdot2^n}{2} $$

Subtract twice the second equation from the first to eliminate the $2^n$ terms to get

$$f_n=4f_{n-1}-4f_{n-2},\quad f_0=1,f_1=8$$

This is a generalized Fibonacci sequence, $f_n=af_{n-1}+bf_{n-2}$ with $f(0)=f_0 \ \&\ f(1)=f_1$. It can be solved with the methods described previously by me here. The general solution is

$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n) $$

where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.

In the present case, we have $\alpha=\beta$ and the solution takes the form

$$ {{f}_{n}}=\left( {{f}_{1}}-\tfrac{a\,{{f}_{0}}}{2} \right)\,n{{\alpha }^{n-1}}+\tfrac{a{{f}_{0}}}{2}\,{{\alpha }^{n-1}}=\left[ n{{f}_{1}}-(n-1)\tfrac{a{{f}_{0}}}{2} \right]\,{{\alpha }^{n-1}} $$

Substituting the parameters for the case at hand, we arrive at

$$a_n=(3n+1)2^n$$

This result has been verified numerically.

Cye Waldman
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$$a_n=3\cdot2^n+2a_{n-1}$$

$$=3\cdot2^n+2(3\cdot2^{n-1}+2a_{n-2})$$ $$=3\cdot2^n(1+1)+2^{1+1}a_{n-2}$$

$$=3\cdot2^nr+2^ra_{n-r}$$ for $0\le r\le n$

lab bhattacharjee
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Hint:

Let $a_m=am2^m+b_m$

$$3\cdot2^n=a_n-2a_{n-1}=an2^n+b_n-2\{a(n-1)2^{n-1}+b_{n-1}\}=a2^n+b_n-2_{n-1}$$

Set $a=3$ to find $b_n-2_{n-1}=0$

lab bhattacharjee
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