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Past exam of the year 1 of the computer science course of the University of Oxford

2021

Discrete mathematics, continuous mathematics, and probability

For $n=1$, both sides are $1/3$. Assuming the formula through $n$, adding the next term gives
$$
\begin{aligned}
\sum_{i=1}^{n+1}\frac1{(2i-1)(2i+1)}
&=\frac{n}{2n+1}+\frac1{(2n+1)(2n+3)}\\
&=\frac{2n^2+3n+1}{(2n+1)(2n+3)}
=\frac{n+1}{2n+3}.
\end{aligned}
$$
The last expression is the required formula at $n+1$, completing <mathematical induction>. The <partial fraction decomposition>
$$
\frac1{(2i-1)(2i+1)}=\frac12\left(\frac1{2i-1}-\frac1{2i+1}\right)
$$
provides a second explanation: this is a <telescoping sum>.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 2 / a / ii / Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 2 / a / ii / Solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 2 / a / ii / Solution

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