Question 2 Prove that -14 + 16 64 256 113 =
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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How would I complete the attached question in Python using only Numpy, recursion loops, or basic math functions?
![### Question 2
Prove that
\[
\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{3}
\]
This mathematical problem asks students to demonstrate that the sum of the infinite series, starting with \(\frac{1}{4}\) and with each subsequent term being the previous term divided by 4, converges to \(\frac{1}{3}\).
The series given in the problem is a geometric series where the first term \(a = \frac{1}{4}\) and the common ratio \(r = \frac{1}{4}\). The formula for the sum \(S\) of an infinite geometric series is:
\[
S = \frac{a}{1 - r}
\]
By substituting the given values:
\[
S = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}
\]
Therefore, the series converges to \(\frac{1}{3}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20ef5b89-bdf5-4ebf-bc1c-34f412b810c9%2Fe1b5f1ea-bb88-4273-904c-1b1308ff2cef%2Ftk4aodq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 2
Prove that
\[
\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{3}
\]
This mathematical problem asks students to demonstrate that the sum of the infinite series, starting with \(\frac{1}{4}\) and with each subsequent term being the previous term divided by 4, converges to \(\frac{1}{3}\).
The series given in the problem is a geometric series where the first term \(a = \frac{1}{4}\) and the common ratio \(r = \frac{1}{4}\). The formula for the sum \(S\) of an infinite geometric series is:
\[
S = \frac{a}{1 - r}
\]
By substituting the given values:
\[
S = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}
\]
Therefore, the series converges to \(\frac{1}{3}\).
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