Use the formula S​ =

n squaredn2

to find the sum of 1​ + 3​ + 5​ + ...​ +

461461.

​(Hint: To find​ n, add 1 to the last term and divide by​ 2.)

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Part 1

1​ + 3​ + 5​ + ...​ + 461

​= enter your response here

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 \(n\).

Discover how to find the sum of the series 1 + 3 + 5 + ... + 461 using the formula S = n^2. Step-by-step solution provided with calculations and explanations. Suitable for Grades 6-8 students learning about series summation and odd numbers.

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