Problem 7 - Thomas Calculus

Question

Each of Exercises \(7-12\) gives the first term or two of a sequence along with a recursion formula for the remaining terms. Write out the first ten terms of the sequence. $$ a_{1}=1, \quad a_{n+1}=a_{n}+\left(1 / 2^{n}\right) $$

Step-by-Step Solution

Answer

The first ten terms are: 1, 1.5, 1.75, 1.875, 1.9375, 1.96875, 1.984375, 1.9921875, 1.99609375, 1.998046875.

Step 1: Understand the Given Information

We have the first term of the sequence, which is \( a_1 = 1 \), and the recursive formula for the remaining terms: \( a_{n+1} = a_n + \frac{1}{2^n} \). Our goal is to find the first ten terms of this sequence.

Step 2: Calculate the Second Term

Using the recursive formula, calculate \( a_2 \). Given \( a_1 = 1 \), the formula is \( a_2 = a_1 + \frac{1}{2^1} = 1 + \frac{1}{2} = 1.5 \).

Step 3: Calculate the Third Term

Now calculate \( a_3 \) using the formula and the previous term, \( a_2 = 1.5 \). \( a_3 = a_2 + \frac{1}{2^2} = 1.5 + \frac{1}{4} = 1.75 \).

Step 4: Calculate the Fourth Term

Using the formula with \( a_3 = 1.75 \), find \( a_4 \). \( a_4 = a_3 + \frac{1}{2^3} = 1.75 + \frac{1}{8} = 1.875 \).

Step 5: Calculate the Fifth Term

Now calculate \( a_5 \) using \( a_4 = 1.875 \). \( a_5 = a_4 + \frac{1}{2^4} = 1.875 + \frac{1}{16} = 1.9375 \).

Step 6: Calculate the Sixth Term

Calculate \( a_6 \) using \( a_5 = 1.9375 \). \( a_6 = a_5 + \frac{1}{2^5} = 1.9375 + \frac{1}{32} = 1.96875 \).

Step 7: Calculate the Seventh Term

Calculate \( a_7 \) using \( a_6 = 1.96875 \). \( a_7 = a_6 + \frac{1}{2^6} = 1.96875 + \frac{1}{64} = 1.984375 \).

Step 8: Calculate the Eighth Term

Calculate \( a_8 \) using \( a_7 = 1.984375 \). \( a_8 = a_7 + \frac{1}{2^7} = 1.984375 + \frac{1}{128} = 1.9921875 \).

Step 9: Calculate the Ninth Term

Calculate \( a_9 \) using \( a_8 = 1.9921875 \). \( a_9 = a_8 + \frac{1}{2^8} = 1.9921875 + \frac{1}{256} = 1.99609375 \).

Step 10: Calculate the Tenth Term

Finally, calculate \( a_{10} \) using \( a_9 = 1.99609375 \). \( a_{10} = a_9 + \frac{1}{2^9} = 1.99609375 + \frac{1}{512} = 1.998046875 \).

Key Concepts

Recursive Formula

A recursive formula is like a recipe for a sequence. It tells you how to find each term in a sequence using the previous ones.
In the given exercise, the recursive formula is written as:
\( a_{n+1} = a_n + \frac{1}{2^n} \).
This formula specifies how you can calculate each term based on the term before it.
By starting from a known starting point, which in this case is \( a_1 = 1 \), you can continue finding new terms.
To solve the exercise, begin with the initial value and apply the formula repeatedly to build the entire sequence step-by-step.
Recursive formulas are powerful because they reduce the dependency on formula memorization and allow sequences to be understood dynamically.
They are essential for finding a large number of sequence terms when only a few starting terms are known.

Sequence Terms

Sequence terms are the individual elements that make up a sequence. In this exercise, each term is calculated using the recursive formula.
The sequence is a list of numbers where each number is derived from the previous numbers by applying the formula.
The first term given is \( a_1 = 1 \).
Starting from here, using the recursive formula, we calculate the following terms:

• Second term: \( a_2 = 1.5 \)
• Third term: \( a_3 = 1.75 \)
• Fourth term: \( a_4 = 1.875 \)
• Fifth term: \( a_5 = 1.9375 \)
• Sixth term: \( a_6 = 1.96875 \)
• Seventh term: \( a_7 = 1.984375 \)
• Eighth term: \( a_8 = 1.9921875 \)
• Ninth term: \( a_9 = 1.99609375 \)
• Tenth term: \( a_{10} = 1.998046875 \)

This list shows how easily each term relates back to the previous ones, illustrating how sequences unfold naturally from initial values.

Mathematical Induction

Mathematical induction is a proof technique used to show that a statement holds for all natural numbers.
Although not directly required in this specific exercise, understanding induction helps to see why recursion is reliable.
Imagine you have shown that a statement is true for an initial case, like \( n = 1 \), and shown that if it's true for \( n \), it's true for \( n+1 \).
This step-by-step validation, akin to dominoes falling, ensures the reliability of results from recursive formulas.
In sequences, you could use induction to prove that a generated formula, when applied indefinitely, will consistently yield accurate results.
This method reassures us that the recursive approach continuously applies accurately, making it a staple technique in mathematics to verify series results.
So while in the exercise we compute the terms through a formula, inductive reasoning is often in the backdrop, ensuring correctness across the steps.