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 \(n\)?
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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 \(8T+1\).
Frequently asked questions
What is the correct answer to this question?
\(8T+1\) is an odd perfect square
Why is this the correct answer?
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 \(8T+1\).
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Sum of first n natural numbers.
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