2026-09-10T00:15:27.642292 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Use the sequence \(4,\ 11,\ 22,\ 37,\ \ldots\) to...
Question 1 Report
Use the sequence \(4,\ 11,\ 22,\ 37,\ \ldots\) to answer all three parts. The sequence is quadratic.
(a) Work out the next term.
(b) Work out an expression for the \(n\)th term.
(c) Work out the 15th term.
Answer Details
A quadratic sequence has constant second differences. Start by finding the differences between the given terms.
| Terms | First differences | Second differences |
|---|---|---|
| \(4,\ 11,\ 22,\ 37\) | \(7,\ 11,\ 15\) | \(4,\ 4\) |
- Since the second difference is 4, the next first difference is \(15+4=19\). Therefore the next term is:
\[37+19=56\]
The next term is \(56\). [B1] - For a quadratic expression \(an^2+bn+c\), the constant second difference is \(2a\). Since \(2a=4\), \(a=2\), so begin with \(2n^2\). [M1]
The values of \(2n^2\) for \(n=1,2,3,4\) are \(2,8,18,32\). Subtract these from the sequence terms:
\[4-2=2,\quad11-8=3,\quad22-18=4,\quad37-32=5\]
The remainders are \(2,3,4,5\), which follow the rule \(n+1\). [M1]
Hence:
\[u_n=2n^2+n+1\]
[A1] - For the 15th term:
\[2(15)^2+15+1=2(225)+16=450+16=466\]
The 15th term is \(466\). [B1]
Examination reminder: a constant second difference identifies a quadratic sequence; divide that second difference by 2 to find the coefficient of \(n^2\).
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