To solve the problem of determining how many bit strings with length not exceeding $n$ consist entirely of 1s, excluding the empty string, let's break it down.

Bit Strings of 1s

A bit string that consists entirely of 1s is a sequence where every element is $1$. We are asked to find how many such bit strings there are with lengths ranging from 1 to $n$.

The possible lengths for the bit strings are 1, 2, 3, ..., up to $n$. So, the bit strings are:

• For length 1: $1$
• For length 2: $11$
• For length 3: $111$
• $\dots$
• For length $n$: $111...1$ (with $n$ 1s)

Total Count of Bit Strings

The number of bit strings consisting entirely of 1s is equal to the number of possible lengths. Since the lengths range from 1 to $n$, there are exactly $n$ such bit strings.

Answer:

The number of bit strings with length not exceeding $n$, consisting entirely of 1s, is $n$.

Do you have any questions, or would you like more details?

Here are 5 related questions to expand your understanding:

1. How many bit strings of length exactly $n$ consist entirely of 1s?
2. What is the total number of bit strings of length $n$, regardless of content?
3. How would the count change if we included the empty string?
4. What if we counted bit strings consisting entirely of 0s instead?
5. How many bit strings of length not exceeding $n$ contain exactly $k$ 1s?

Tip: Bit strings can be useful in combinatorics and computer science when working with binary operations or coding theory.