Prove that the arithmetic sequence with first term \(\frac{1}{2}\) and common difference \(\frac{1}{4}\) contains all natural numbers?
Answer (2)
The arithmetic sequence with a first term of 2 1 and a common difference of 4 1 does not contain all natural numbers. It generates specific natural numbers based on the formula derived from the sequence. Hence, many natural numbers are excluded from this sequence.
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To determine if the arithmetic sequence with a first term of 2 1 and a common difference of 4 1 contains all natural numbers, let's analyze the sequence.
An arithmetic sequence is defined by its first term a 1 and common difference d . The general term of an arithmetic sequence (the nth term) can be expressed as:
a n = a 1 + ( n − 1 ) ⋅ d
In this case, the first term a 1 is 2 1 and the common difference d is 4 1 . Thus, the nth term is given by:
a n = 2 1 + ( n − 1 ) ⋅ 4 1
Simplifying this expression, we get:
a n = 2 1 + 4 n − 1 = 4 2 + 4 n − 1 = 4 n + 1
To see if this sequence contains all natural numbers, we need to determine if this expression can yield every natural number for some integer value of n .
A natural number is an integer greater than or equal to 1. Let's set a n = k , where k is a natural number:
4 n + 1 = k
Solving for n , we multiply both sides by 4:
n + 1 = 4 k
Subtracting 1 from both sides gives:
n = 4 k − 1
For n to be a non-negative integer, 4 k − 1 must also be non-negative, which implies k ≥ 1 . This shows that for every natural number k , there is a corresponding n = 4 k − 1 that generates a n = k .
Therefore, every natural number can be represented by some term in this arithmetic sequence, and thus the sequence contains all natural numbers.
