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Which condition is necessary and sufficient for a positive integer \(T\) to be the sum of the first \(n\) natural numbers for some positive integer \(n\)?

Answer and explanation

Correct answer: \(8T+1\) is an odd perfect square

If \(T=\frac{n(n+1)}{2}\), then \(8T+1=4n(n+1)+1=(2n+1)^2\), which is an odd perfect square. The converse is also true. \(4T+1\) has no such guaranteed form. Exam tip: test \(8T+1\).

Related tags

Triangular NumbersSum Of Natural NumbersSequences And ProgressionsNumber PropertiesPerfect SquaresSum Of First N Natural NumbersMathematicsClass 9 Mcq

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