SPS SPS FM 2020 December — Question 12 7 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2020
SessionDecember
Marks7
TopicBinomial Theorem (positive integer n)
TypeCoefficients in arithmetic/geometric progression
DifficultyStandard +0.3 This is a straightforward binomial expansion question requiring standard techniques: writing out terms with binomial coefficients, setting up an arithmetic sequence condition, and solving a cubic that factors nicely. The algebra is routine and the cubic conveniently factors with n as a common factor, making it easier than average but not trivial due to the multi-step nature and 7 total marks.
Spec1.04a Binomial expansion: (a+b)^n for positive integer n1.04h Arithmetic sequences: nth term and sum formulae

Consider the binomial expansion of \(\left(1 + \frac{x}{5}\right)^n\) in ascending powers of \(x\), where \(n\) is a positive integer.
  1. Write down the first four terms of the expansion, giving the coefficients as polynomials in \(n\). [1]
The coefficients of the second, third and fourth terms of the expansion are consecutive terms of an arithmetic sequence.
  1. Show that \(n^3 - 33n^2 + 182n = 0\). [3]
  2. Hence find the possible values of \(n\) and the corresponding values of the common difference. [3]

Consider the binomial expansion of $\left(1 + \frac{x}{5}\right)^n$ in ascending powers of $x$, where $n$ is a positive integer.
\begin{enumerate}[label=(\roman*)]
\item Write down the first four terms of the expansion, giving the coefficients as polynomials in $n$. [1]
\end{enumerate}
The coefficients of the second, third and fourth terms of the expansion are consecutive terms of an arithmetic sequence.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $n^3 - 33n^2 + 182n = 0$. [3]
\item Hence find the possible values of $n$ and the corresponding values of the common difference. [3]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM 2020 Q12 [7]}}