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 defined by the first term 2 1 and a common difference of 4 1 can generate any natural number. By setting the general term equal to any natural number k , we find that there exists a corresponding natural number n for each k . Hence, the sequence includes all natural numbers.
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To determine whether the arithmetic sequence with the first term 2 1 and a common difference of 4 1 contains all natural numbers, we need to explore the expression for the nth term of this sequence.
The nth term of an arithmetic sequence can be expressed as:
a n = a 1 + ( n − 1 ) d
where a 1 is the first term and d is the common difference.
For this sequence, a 1 = 2 1 and d = 4 1 . Thus, the nth term is:
a n = 2 1 + ( n − 1 ) ⋅ 4 1
Simplifying this, we have:
a n = 2 1 + 4 n − 1
a n = 4 2 + 4 n − 1
a n = 4 2 + n − 1
a n = 4 n + 1
To find out if a natural number x can be part of this sequence, we need to set a n = x and solve for n :
x = 4 n + 1
Multiplying both sides by 4 gives:
4 x = n + 1
Therefore:
n = 4 x − 1
For n to be a natural number, 4 x − 1 must be a positive integer, which is true for all natural numbers x since x starts from 1 and increases.
Therefore, every natural number x can be represented by some value of n in this sequence, meaning that this arithmetic sequence does indeed contain all natural numbers.
