STEP I 2005 7

(i)

We have:

\begin{align*}\prod_{r = 1}^n \left(\frac {r + 1} r\right) &= \frac {\cancel 2} 1 \times \frac {\cancel 3} {\cancel 2} \times \frac {\cancel 4} {\cancel 3} \ldots \times \frac {\cancel n} {\cancel {n - 1}} \times \frac {n + 1} {\cancel n} \\ & = n + 1\end{align*}

(ii)

We have:

\begin{align*}\prod_{r = 2}^n \left(\frac {r^2 - 1} {r^2}\right) & = \prod_{r = 2}^n \left(\frac {(r - 1)(r + 1)} {r^2}\right) \\ & = \prod_{r = 2}^n \left(\frac {r - 1} r\right) \prod_{r = 2}^n \left(\frac {r + 1} r\right)\end{align*}

Using part (i):

\displaystyle \prod_{r = 2}^n \left(\frac {r + 1} r\right) = \frac 1 {1 + 1} \prod_{r = 1}^n \left(\frac {r + 1} r\right) = \frac {n + 1} 2

We also have:

\begin{align*} \prod_{r = 2}^n \left(\frac {r - 1} r\right) & = \frac 1 {\cancel 2} \times \frac {\cancel 2} {\cancel 3} \times \frac {\cancel 3} {\cancel 4} \ldots \frac {\cancel {n - 2}} {\cancel{n - 1}} \times \frac {\cancel{n - 1}} n \\ & = \frac 1 n\end{align*}

So:

\displaystyle \prod_{r = 2}^n \left(\frac {r^2 - 1} {r^2}\right) = \frac {n + 1} {2n} = \frac 1 2\left(1 + \frac 1 n\right)

Alternative approaches for these two parts include taking logarithms (then dealing with a telescoping sum) or using the fact that \prod_{r = 1}^n r = n! and then cancelling.

(iii)

For brevity, write the desired product as P.

We have:

\begin{align*}\cos \frac {2 \pi} n + \sin \frac {2 \pi} n \cot \frac {(2r - 1) \pi} n & = \cos \frac {2 \pi} n + \frac {\sin \frac {2 \pi} n \cos \frac {(2r - 1) \pi} n} {\sin \frac {(2r - 1) \pi} n} \\ & = \frac {\cos \frac {2 \pi} n \sin \frac {(2r - 1) \pi} n +\sin \frac {2 \pi} n \cos \frac {(2r - 1) \pi} n} {\sin \frac {(2r - 1) \pi} n} \\ & = \frac {\sin \left(\frac {(2r - 1) \pi} n + \frac {2 \pi} n\right)} {\sin \frac {(2r - 1) \pi} n} \\ & = \frac {\sin \frac {(2r + 1) \pi} n} {\sin \frac {(2r - 1) \pi} n} \end{align*}

We have:

\begin{align*}P &= \prod_{r = 1}^n \left(\frac {\sin \frac {(2r + 1) \pi} n} {\sin \frac {(2r - 1) \pi} n}\right) \\ & = \frac {\cancel {\sin \frac {3 \pi} n}} {\sin \frac \pi n} \times \frac {\cancel{\sin \frac {5 \pi} n}} {\cancel{\sin \frac {3 \pi} n}} \times \ldots \frac {\cancel{\sin \frac {(2n - 1) \pi} n}} {\cancel{\sin \frac {(2n - 3) \pi} n}} \times \frac {\sin \frac {(2n + 1) \pi} n} {\cancel{\sin \frac {(2n - 1) \pi} n}} \\ & = \frac {\sin \frac {(2n + 1) \pi} n} {\sin \frac \pi n} \\ & = \frac {\sin \left(2 \pi + \frac \pi n\right)} {\sin \frac \pi n} \\ & = \frac {\sin \frac \pi n} {\sin \frac \pi n} \\ & = 1\end{align*}

(with n even preventing possibly zero denominators if (2r - 1)/n is ever an integer)

1 Like

:clap: