What is the (n)th term of the sequence (1,3,7,13,21,\ldots)?
Answer and explanation
Correct answer: \(n^2-n+1\)
The consecutive differences are \(2,4,6,8,\ldots\), so the added differences up to the \(n\)th term are \(2(k-1)\). Hence, \(a_n=1+2(1+2+\cdots +(n-1))=1+n(n-1)=n^2-n+1\). The close distractor \(n^2+n-1\) gives \(5\) when \(n=2\), whereas the second term is \(3\). Exam tip: if first differences increase by a constant amount, the general term is usually quadratic in \(n\).
Frequently asked questions
What is the correct answer to this question?
\(n^2-n+1\)
Why is this the correct answer?
The consecutive differences are \(2,4,6,8,\ldots\), so the added differences up to the \(n\)th term are \(2(k-1)\). Hence, \(a_n=1+2(1+2+\cdots +(n-1))=1+n(n-1)=n^2-n+1\). The close distractor \(n^2+n-1\) gives \(5\) when \(n=2\), whereas the second term is \(3\). Exam tip: if first differences increase by a constant amount, the general term is usually quadratic in \(n\).
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.
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