Olympiad Maths Prep

Track / Stage 5 / 133 of 400 #733 of 2000

Problem 733

AIME late
Algebra Difficulty 5.3 Find the answer

2. In the sequence {an}\left\{a_{n}\right\}, it is known that an+2=3an+1−2an,a1=1,a2=3a_{n+2}=3 a_{n+1}-2 a_{n}, a_{1}=1, a_{2}=3, then the general term formula of the sequence {an}\left\{a_{n}\right\} is an=a_{n}=

Official solution

an+2−an+1=2(an+1−an)⇒an+1−an=2n⇒an−a1=2+22+⋯+2n−1=2n−2⇒an=2n−1(n∈N+) \begin{array}{l} a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_{n}\right) \Rightarrow a_{n+1}-a_{n}=2^{n} \Rightarrow a_{n}-a_{1}=2+2^{2}+\cdots+2^{n-1} \\ =2^{n}-2 \Rightarrow a_{n}=2^{n}-1\left(n \in \mathbf{N}_{+}\right) \end{array}

The translation is as follows:

an+2−an+1=2(an+1−an)⇒an+1−an=2n⇒an−a1=2+22+⋯+2n−1=2n−2⇒an=2n−1(n∈N+) \begin{array}{l} a_{n+2}-a_{n+1}=2\left(a_{n+1}-a_{n}\right) \Rightarrow a_{n+1}-a_{n}=2^{n} \Rightarrow a_{n}-a_{1}=2+2^{2}+\cdots+2^{n-1} \\ =2^{n}-2 \Rightarrow a_{n}=2^{n}-1\left(n \in \mathbf{N}_{+}\right) \end{array}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.