Exercise 355Calculus › Single-variable calculus › SequencesL1Added 2 Nov 2026

A linear recurrence of order two

#linear recurrence#characteristic equation

Statement

Find the general term of $(u_n)$ with $u_0=3$ , $u_1=5$ and $u_{n+2}=3u_{n+1}-2u_n$ .

Hint

Look for solutions $u_n=r^n$ . This leads to the characteristic equation $r^2-3r+2=0$ .

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Characteristic equation. $r^2-3r+2=(r-1)(r-2)=0$ , so $r_1=1$ and $r_2=2$ .

General solution. Both $1^n$ and $2^n$ satisfy the recurrence, and so does every combination $u_n=A+B2^n$ .

Initial conditions. $A+B=3$ and $A+2B=5$ , so $B=2$ and $A=1$ .

Answer. $u_n=1+2^{n+1}$ .