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Class 9MathematicsSequences and Progressionsnth termExpertLevel 57
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If (a_n=n^3), what is the formula for (a_n-a_{n-1})?

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Answer and explanation

Correct answer: \(3n^2-3n+1\)

Here, \(a_{n-1}=(n-1)^3\). Therefore, \(a_n-a_{n-1}=n^3-(n-1)^3=n^3-(n^3-3n^2+3n-1)=3n^2-3n+1\). Hence, option D is correct. Option A is the result of \((n+1)^3-n^3\), not of the given difference. Exam tip: for the previous term, always replace \(n\) with \(n-1\).

Related tags

SequencesProgressionsNth TermFinite DifferencesCubic SequenceClass 9

Frequently asked questions

What is the correct answer to this question?

\(3n^2-3n+1\)

Why is this the correct answer?

Here, \(a_{n-1}=(n-1)^3\). Therefore, \(a_n-a_{n-1}=n^3-(n-1)^3=n^3-(n^3-3n^2+3n-1)=3n^2-3n+1\). Hence, option D is correct. Option A is the result of \((n+1)^3-n^3\), not of the given difference. Exam tip: for the previous term, always replace \(n\) with \(n-1\).

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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