Answer only one of the following two alternatives.

**EITHER**

Show that $\left(n + \frac{1}{2}\right)^3 - \left(n - \frac{1}{2}\right)^3 \equiv 3n^2 + \frac{1}{4}$. [1]

Use this result to prove that $\sum_{n=1}^N n^2 = \frac{1}{6}N(N + 1)(2N + 1)$. [2]

The sums $S$, $T$ and $U$ are defined as follows:
\begin{align}
S &= 1^2 + 2^2 + 3^2 + 4^2 + \ldots + (2N)^2 + (2N + 1)^2, \\
T &= 1^2 + 3^2 + 5^2 + 7^2 + \ldots + (2N - 1)^2 + (2N + 1)^2, \\
U &= 1^2 - 2^2 + 3^2 - 4^2 + \ldots - (2N)^2 + (2N + 1)^2.
\end{align}

Find and simplify expressions in terms of $N$ for each of $S$, $T$ and $U$. [5]

Hence
\begin{enumerate}[label=(\roman*)]
\item describe the behaviour of $\frac{S}{T}$ as $N \to \infty$, [1]
\item prove that if $\frac{S}{U}$ is an integer then $\frac{T}{U}$ is an integer. [3]
\end{enumerate}

**OR**

The curves $C_1$ and $C_2$ have polar equations
$$r = 4\cos\theta \quad \text{and} \quad r = 1 + \cos\theta$$
respectively, where $-\frac{1}{2}\pi \leqslant \theta \leqslant \frac{1}{2}\pi$.

\begin{enumerate}[label=(\roman*)]
\item Show that $C_1$ and $C_2$ meet at the points $A\left(\frac{4}{3}, \alpha\right)$ and $B\left(\frac{4}{3}, -\alpha\right)$, where $\alpha$ is the acute angle such that $\cos\alpha = \frac{1}{3}$. [2]
\item In a single diagram, draw sketch graphs of $C_1$ and $C_2$. [3]
\item Show that the area of the region bounded by the arcs $OA$ and $OB$ of $C_1$, and the arc $AB$ of $C_2$, is
$$4\pi - \frac{1}{3}\sqrt{2} - \frac{13}{2}\alpha.$$ [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2005 Q12 [24]}}