Math Problem Statement

Use the formula S ​=StartFraction n left parenthesis n plus 1 right parenthesis Over 2 EndFraction to find the sum of 1​ + 2​ + 3​ + ...​ + 880. Question content area bottom Part 1 1​ + 2​ + 3​ + ...​ + 880 ​=

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Solution

The formula provided, S=n(n+1)2S = \frac{n(n+1)}{2}, is used to find the sum of the first nn natural numbers.

In this case, the sum of the series 1+2+3++8801 + 2 + 3 + \dots + 880 can be calculated by setting n=880n = 880.

Let's apply the formula:

S=880(880+1)2S = \frac{880(880 + 1)}{2}

S=880×8812S = \frac{880 \times 881}{2}

S=7740802S = \frac{774080}{2}

S=387440S = 387440

So, the sum 1+2+3++8801 + 2 + 3 + \dots + 880 is 387,440387,440.

Would you like further details or have any questions?

Here are 5 related questions:

  1. How does this formula change if the series does not start from 1?
  2. What is the sum of the first 1000 natural numbers?
  3. How would you find the sum of an arithmetic series that has a common difference other than 1?
  4. Can this formula be used for non-consecutive numbers?
  5. How do you derive the formula for the sum of the first nn natural numbers?

Tip: The formula S=n(n+1)2S = \frac{n(n+1)}{2} is a quick way to sum up consecutive integers starting from 1. It's a valuable tool for solving problems efficiently!

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Math Problem Analysis

Mathematical Concepts

Summation of Series
Arithmetic Series

Formulas

Sum of first n natural numbers: S = n(n + 1) / 2

Theorems

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Suitable Grade Level

Grades 6-8