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If (a_n=n^2), what is the formula for (a_n-a_{n-1})?
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Answer and explanation
Correct answer: \(2n-1\)
Given \(a_n=n^2\), the previous term is \(a_{n-1}=(n-1)^2\). Therefore, \(a_n-a_{n-1}=n^2-(n-1)^2=n^2-(n^2-2n+1)=2n-1\). The expression \(2n+1\) is obtained from \((n+1)^2-n^2\), so it is not correct here. Exam tip: To find \(a_{n-1}\), replace every \(n\) by \(n-1\).
Frequently asked questions
What is the correct answer to this question?
\(2n-1\)
Why is this the correct answer?
Given \(a_n=n^2\), the previous term is \(a_{n-1}=(n-1)^2\). Therefore, \(a_n-a_{n-1}=n^2-(n-1)^2=n^2-(n^2-2n+1)=2n-1\). The expression \(2n+1\) is obtained from \((n+1)^2-n^2\), so it is not correct here. Exam tip: To find \(a_{n-1}\), replace every \(n\) by \(n-1\).
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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