A Skip-Counting Sequence

Problem

A sequence {an}\{a_n\} has a1=1a_1 = 1, and for n2n \geqslant 2,

an={an1,n4N,an1+1,n4N.a_n = \begin{cases} a_{n-1}, & \dfrac{n}{4} \in \mathbb{N}^*, \\[2pt] a_{n-1} + 1, & \dfrac{n}{4} \notin \mathbb{N}^*. \end{cases}

Let bmb_m be the number of terms of {an}\{a_n\} lying in [2m,2m+1)\left[2^m, 2^{m+1}\right), for mNm \in \mathbb{N}^*. Determine, with proof, which of the following are true:

1. b3=10b_3 = 10; 2. 3n4an3n+34\dfrac{3n}{4} \leqslant a_n \leqslant \dfrac{3n+3}{4}; 3. k=14nak=6n2+3n\displaystyle\sum_{k=1}^{4n} a_k = 6n^2 + 3n; 4. k=12nbk=43(4n1)\displaystyle\sum_{k=1}^{2n} b_k = \dfrac{4}{3}\left(4^n - 1\right).

Answer

Solution

Difficulty7/10
TopicsArithmetic Progression, sequences, Casework

Whiteboard

Your sketch is saved only in this browser. To share it, export your drawing as an image (whiteboard menu → Export as → PNG), then upload that image in the comments below.

Discussion

Ask questions, share alternate solutions, and use LaTeX freely.

0 comments
Log in to join the discussion.

No comments yet.