To find the sum of the series $1 + 3 + 5 + \dots + 461$, we can use the formula for the sum of the first $n$ odd numbers:

$S = n^2$

Where $n$ is the number of terms in the series.

Step 1: Find $n$

The hint suggests that to find $n$, we need to add 1 to the last term (461) and divide by 2:

$n = \frac{461 + 1}{2} = \frac{462}{2} = 231$

So, $n = 231$.

Step 2: Calculate the sum $S$

Now, use the formula $S = n^2$ to find the sum:

$S = 231^2$

I'll calculate that next.The sum of the series $1 + 3 + 5 + \dots + 461$ is $53361$.

Would you like more details or have any questions? Here are some related questions you might find helpful:

1. How would you find the sum of even numbers in a similar series?
2. What is the general formula for the sum of the first $n$ odd numbers, and why does it work?
3. How can you verify that $n = 231$ is correct for this problem?
4. What is the sum of all odd numbers up to a different value, such as 501?
5. How can this method be adapted for finding the sum of a series that skips terms?

Tip: Remember that the sum of the first $n$ odd numbers is always $n^2$, which makes finding such sums straightforward if you can determine $n$.