Math Problem Statement

If we assume a natural number and prove n>2, the expression 1+2+3+...+n is a composite number.

Solution

The given problem asks us to prove that the sum of the first nn natural numbers is a composite number if n>2n > 2.

The sum of the first nn natural numbers is given by the formula: S=1+2+3++n=n(n+1)2S = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}

We need to show that SS is a composite number for n>2n > 2.

Step-by-step proof:

  1. Expression for the sum: The sum of the first nn natural numbers is S=n(n+1)2S = \frac{n(n+1)}{2}.

  2. Factor properties:

    • The product n(n+1)n(n+1) is always even because one of nn or n+1n+1 is even.
    • Dividing by 2 ensures that SS is an integer.
  3. Analyze for small values of nn: Let's check the expression for small values of nn:

    • For n=1n = 1: S=1(1+1)2=1S = \frac{1(1+1)}{2} = 1 (not composite, but we are interested in n>2n > 2).
    • For n=2n = 2: S=2(2+1)2=3S = \frac{2(2+1)}{2} = 3 (not composite).
    • For n=3n = 3: S=3(3+1)2=6S = \frac{3(3+1)}{2} = 6 (composite).
    • For n=4n = 4: S=4(4+1)2=10S = \frac{4(4+1)}{2} = 10 (composite).
    • For n=5n = 5: S=5(5+1)2=15S = \frac{5(5+1)}{2} = 15 (composite).

    For n>2n > 2, it appears that SS is composite. We now generalize this observation.

  4. Generalization: The sum S=n(n+1)2S = \frac{n(n+1)}{2} is the product of two integers, nn and n+12\frac{n+1}{2}, when nn is odd, or n2\frac{n}{2} and n+1n+1, when nn is even. In either case, as long as n>2n > 2, these integers will be greater than 1, which means SS is a product of two integers greater than 1, i.e., SS is composite.

Conclusion:

For n>2n > 2, the sum S=n(n+1)2S = \frac{n(n+1)}{2} is always a composite number, as it is the product of two integers both greater than 1.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

  1. What happens if n=2n = 2? Can we prove this special case differently?
  2. How can we determine whether a number is composite or prime?
  3. How does the sum formula change if we include only even or odd numbers?
  4. What if we consider other patterns of natural numbers, like squares or cubes? Will their sums behave similarly?
  5. Can this method be extended to sums of arithmetic sequences with different starting points and common differences?

Tip: When trying to prove that a number is composite, factorization is key. Try to express the number as a product of two integers greater than 1.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Sum of Natural Numbers
Composite Numbers

Formulas

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

Theorems

Composite Number Definition
Basic Properties of Even and Odd Numbers

Suitable Grade Level

Grades 9-11