Problem 17 - Algebra 2 and Trigonometry

Question

The sum of the infinite series \(1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n}}\) is \(2 .\) Find values of \(n\) such that \(2-a_{n}<\frac{226-1}{2^{25}}\)

Step-by-Step Solution

Answer

The value of \( n \) that satisfies the condition is 18.

Step 1: Recognize the Series Type

The series in question is a geometric series, where each term after the first is found by multiplying the previous term by a constant, which in this case is \( \frac{1}{2} \).

Step 2: Geometric Series Formula

For a geometric series with first term \(a = 1\) and common ratio \(r = \frac{1}{2} \), the sum \(S\) of the infinite series is given by the formula \( S = \frac{a}{1 - r} \). Therefore, \( S = \frac{1}{1 - \frac{1}{2}} = 2 \).

Step 3: Partial Sum of the Series

The sum of the first \( n \) terms of the geometric series can be calculated by \( S_n = a \frac{1-r^n}{1-r} \). Using \( a = 1 \) and \( r = \frac{1}{2} \), it becomes \( S_n = \frac{1 - (\frac{1}{2})^n}{1 - \frac{1}{2}} = 2(1 - \frac{1}{2^n}) \).

Step 4: Express the Problem Condition

We need to solve for \( n \) in the inequality \( 2 - a_n < \frac{226 - 1}{2^{25}} \), where \( a_n = S_n = 2(1 - \frac{1}{2^n}) \).

Step 5: Simplify the Expression

Substitute \( a_n \) into the inequality: \( 2 - 2(1 - \frac{1}{2^n}) < \frac{225}{2^{25}} \), simplifying to \( \frac{2}{2^n} < \frac{225}{2^{25}} \).

Step 6: Solve the Inequality

Bring the terms together: \( 2^{25 - n} < 225 \). Calculate logarithmically or estimate powers of 2 to find the largest integer \( n \) such that \( 2^{25 - n} < 225 \).

Step 7: Calculate and Verify

Recall that \( 2^7 = 128 \) and \( 2^8 = 256 \). Since \( 2^7 < 225 < 2^8 \), we need \( 25 - n = 7 \); thus, \( n = 18 \).

Key Concepts

Infinite Series

An infinite series is a sequence of numbers that go on forever. Specifically, it is a sum where there are infinitely many terms.
In our case, the series under discussion is: \[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \]Each term is generated by dividing the previous term by 2, and this continues endlessly.
The series converges, meaning it approaches a particular value, which here happens to be 2.
This sum of an infinite series, when possible, often simplifies complex mathematical calculations such as repetitive processes and phenomena transformation.

Partial Sum

A partial sum is the sum of the first few terms of a series. For a geometric series like ours, it captures only a finite number of the terms.
This is useful to estimate the value of an infinite series without adding infinitely many terms.
For the given series, the formula for the partial sum of the first \( n \) terms is given by:\[ S_n = a \frac{1-r^n}{1-r} \] where \( a = 1 \) and \( r = \frac{1}{2} \). So, \( S_n = 2\left(1 - \frac{1}{2^n}\right) \).
This helps you see how close a specific number of terms gets you to the total value of the infinite series.
By solving it, you can understand how quickly or slowly the series converges."},{

Geometric Series Formula

The geometric series formula is a mathematical expression that allows us to find the sum of a geometric series. Such series have a constant ratio between consecutive terms.
In our example, the first term \(a\) is 1, and the common ratio \(r\) is \(\frac{1}{2}\).
To find the sum \(S\) of an infinite geometric series, we use:\[ S = \frac{a}{1 - r} \]Substituting the known values, we get:\[ S = \frac{1}{1 - \frac{1}{2}} = 2 \]This formula only works when the absolute value of the ratio \(|r|\) is less than 1. When \(|r| < 1\), the series converges, which means it approaches a specific number.

Series Inequality

Inequalities in series involve expressions showing some form of less than or greater than relation concerning series terms and sums.
The problem condition is given by the inequality \( 2 - a_n < \frac{225}{2^{25}} \).
You substitute \( a_n \) with the partial sum formula and get the inequality:\[ \frac{2}{2^n} < \frac{225}{2^{25}} \]The task then is to solve for \( n \) which satisfies the inequality.
This type of calculation often involves estimating or using logarithms to find the largest possible integer value of \( n \). Here, we determined \( n = 18 \) using such methods, showing how the series quickly approaches its limit.