If (a_n=5n^2-2n+1), what is (a_{n+1}-a_n)?
Answer and explanation
Correct answer: \(10n+3\)
Given \(a_n=5n^2-2n+1\), substitute \(n+1\) for \(n\): \(a_{n+1}=5(n+1)^2-2(n+1)+1=5n^2+8n+4\). Therefore, \(a_{n+1}-a_n=(5n^2+8n+4)-(5n^2-2n+1)=10n+3\). The distractor \(10n-2\) results from missing the \(10n\) term produced while expanding \((n+1)^2\). Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).
Frequently asked questions
What is the correct answer to this question?
\(10n+3\)
Why is this the correct answer?
Given \(a_n=5n^2-2n+1\), substitute \(n+1\) for \(n\): \(a_{n+1}=5(n+1)^2-2(n+1)+1=5n^2+8n+4\). Therefore, \(a_{n+1}-a_n=(5n^2+8n+4)-(5n^2-2n+1)=10n+3\). The distractor \(10n-2\) results from missing the \(10n\) term produced while expanding \((n+1)^2\). Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).
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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