Question

22. a) For $n \in Z^+$, let $a_n$ count the number of palindromes of $2n$. Then $a_{n+1} = 2a_n$, $n \ge 1$, $a_1 = 2$. Solve this first-order recurrence relation for $a_n$. b) For $n \in Z^+$, let $b_n$ count the number of palindromes of $2n-1$. Set up and solve a first-order recurrence relation for $b_n$.

Name: 22 a for n in z let an count the number of palindromes of 2n then an1 2an n ge 1 a1 2 solve this first order recurrence relation for an b for n in z let bn count the number of palindro 18922
Uploaded: 2024-04-27T10:06:23-08:00
Duration: 4 min 6 s
Channel: Adi S
Description: 22 a for n in z let an count the number of palindromes of 2n then an1 2an n ge 1 a1 2 solve this first order recurrence relation for an b for n in z let bn count the number of palindro 18922

          22. a) For $n \in Z^+$, let $a_n$ count the number of palindromes of $2n$. Then $a_{n+1} = 2a_n$, $n \ge 1$, $a_1 = 2$. Solve this first-order recurrence relation for $a_n$.
b) For $n \in Z^+$, let $b_n$ count the number of palindromes of $2n-1$. Set up and solve a first-order recurrence relation for $b_n$.

22 a for n in z let an count the number of palindromes of 2n then an1 2an n ge 1 a1 2 solve this first order recurrence relation for an b for n in z let bn count the number of palindro 18922

Added by Aaron J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

04/27/2024

Step 1

This is a geometric sequence with a common ratio of 2. We can write $a_n = a_1 \cdot r^{n-1}$, where $a_1 = 2$ and $r = 2$. Thus, $a_n = 2 \cdot 2^{n-1} = 2^n$. Show more…

Show all steps

Thanks for your feedback!

22. a) For $n \in Z^+$, let $a_n$ count the number of palindromes of $2n$. Then $a_{n+1} = 2a_n$, $n \ge 1$, $a_1 = 2$. Solve this first-order recurrence relation for $a_n$. b) For $n \in Z^+$, let $b_n$ count the number of palindromes of $2n-1$. Set up and solve a first-order recurrence relation for $b_n$.