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Under which condition is the sum of the first n natural numbers a multiple of n?

Answer and explanation

Correct answer: When n is odd

The sum is \(S_n=\frac{n(n+1)}{2}\). Dividing by n gives \(\frac{n+1}{2}\), which is an integer only when n is odd. Even n leaves a half-integer. Exam tip: test divisibility by cancelling n first.

Related tags

Natural NumbersSum FormulaDivisibilityNumber PropertiesSequences And ProgressionsSum Of First N Natural NumbersMathematicsClass 9 Mcq

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