Recursion

Art of Problem Solving

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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Sid-darth-vater

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Sid-darth-vater

#1 h Apr 20, 2025, 3:02 AM

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Help, I can't characterize ts and I dunno what to do

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#2 h Apr 20, 2025, 5:14 AM

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well by testing small cases i got the general form to be Click to reveal hidden text, which can probably be proven by induction

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Sid-darth-vater

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Sid-darth-vater

#3 h Apr 20, 2025, 3:40 PM

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Ahh yeah, it works out. tyty. also, can you kinda explain how you thought of the numerator part? like I had figured out $a_2 = \frac{7}{24}, a_3 = \frac{13}{96},$ and $a_4 = \frac{25}{384}$ but I couldn't find the $3 \cdot 2^{n-1} + 1$ portion.

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#4 h Apr 20, 2025, 4:22 PM

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$a_{1}=\frac{4}{6},a_{2}=\frac{7}{24},a_{3}=\frac{13}{96},a_{4}=\frac{25}{384},a_{5}=\frac{49}{1536},a_{6}=\frac{97}{6144},\cdots$

$a_{n}=\frac{3 \cdot 2^{n-1} +1}{3 \cdot 2^{2n-1}}$ .

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#5 h Apr 20, 2025, 4:55 PM

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Sid-darth-vater wrote:

Ahh yeah, it works out. tyty. also, can you kinda explain how you thought of the numerator part? like I had figured out $a_2 = \frac{7}{24}, a_3 = \frac{13}{96},$ and $a_4 = \frac{25}{384}$ but I couldn't find the $3 \cdot 2^{n-1} + 1$ portion.

well I noticed that the differences between consecutive numerators was 3, 6, 12, etc.
and so I got a geometric series which resulted in that

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Sid-darth-vater

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Sid-darth-vater

#6 h Apr 20, 2025, 5:52 PM

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mmmm i see, thanks!

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#7 h Apr 20, 2025, 5:59 PM

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$a_{k}=\frac{3 \cdot 2^{k-1} +1}{3 \cdot 2^{2k-1}}=2^{-k}+\frac{1}{3} \cdot 2^{1-2k}$ .

$S=\lim_{n \to \infty} \sum_{k=1}^{n}a_{k}=\lim_{n \to \infty} \sum_{k=1}^{n}\left[2^{-k}+\frac{1}{3} \cdot 2^{1-2k}\right]$ ,
$S=\lim_{n \to \infty} \left[1-2^{-n}+\frac{1}{3}(\frac{2}{3}-\frac{2^{1-2n}}{2})\right]=1+\frac{2}{9}=\frac{11}{9}$ .

This post has been edited 1 time. Last edited by vanstraelen, Apr 20, 2025, 6:00 PM

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