Prove that the arithmetic sequence with first term [tex]\( \frac{1}{2} \)[/tex] and common difference [tex]\( \frac{1}{4} \)[/tex] contains all natural numbers.
Answer (2)
The arithmetic sequence with first term 2 1 and common difference 4 1 contains all natural numbers because its n-th term can be expressed as a n = 4 n + 1 . By choosing suitable values for n , any natural number can be produced. This formula confirms that there is no natural number that cannot be generated by the sequence.
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To prove that the arithmetic sequence with a first term 2 1 and a common difference 4 1 contains all natural numbers, we need to express the general term of this sequence and then demonstrate that for every natural number n , there exists an integer k such that the term a k is equal to n .
The general term a k of an arithmetic sequence is given by the formula:
a k = a 1 + ( k − 1 ) ⋅ d
where a 1 is the first term and d is the common difference. For this sequence:
a 1 = 2 1
d = 4 1
So the formula for the general term becomes:
a k = 2 1 + ( k − 1 ) ⋅ 4 1
Simplifying this, we have:
a k = 2 1 + 4 k − 1
a k = 4 2 + 4 k − 1
a k = 4 2 + k − 1
a k = 4 k + 1
To determine if a natural number n can be a term in the sequence, set a k = n :
4 k + 1 = n
Solving for k :
k + 1 = 4 n
k = 4 n − 1
Here, k is an integer when 4 n − 1 is an integer, which is true for all natural numbers n . Therefore, each natural number n corresponds to a term a k in the sequence, specifically when k = 4 n − 1 .
Thus, the given arithmetic sequence contains all natural numbers.
