Math Problem Statement

Use the formula S​ =

n squaredn2

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

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

Question content area bottom

Part 1

1​ + 3​ + 5​ + ...​ + 883

​=

Solution

To find the sum of the series 1+3+5++8831 + 3 + 5 + \dots + 883, we can again use the formula for the sum of the first nn odd numbers:

S=n2S = n^2

Step 1: Find nn

The hint suggests that to find nn, we need to add 1 to the last term (883) and divide by 2:

n=883+12=8842=442n = \frac{883 + 1}{2} = \frac{884}{2} = 442

So, n=442n = 442.

Step 2: Calculate the sum SS

Now, use the formula S=n2S = n^2 to find the sum:

S=4422S = 442^2

I'll calculate that next.The sum of the series 1+3+5++8831 + 3 + 5 + \dots + 883 is 195364195364.

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

  1. How would the sum change if the last term were 885 instead of 883?
  2. How does the pattern of odd numbers relate to perfect squares?
  3. Can you derive the formula for the sum of odd numbers from first principles?
  4. How would you calculate the sum of even numbers up to a specific term, like 882?
  5. What is the sum of the first kk even numbers, and how does that compare to the sum of the first kk odd numbers?

Tip: For large series, always double-check the value of nn by verifying that n×21n \times 2 - 1 gives the correct last term in the sequence.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Series
Summation of Odd Numbers
Perfect Squares

Formulas

Sum of first n odd numbers: S = n^2

Theorems

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Suitable Grade Level

Grades 7-9