- Prove that the sum of all the integers between \(m\) and \(n\) inclusive (\(m, n \in \mathbb{Z}_+\), \(n > m\)) is \(\tfrac{1}{2}(m+n)(n-m+1)\).

Find the sum of all the integers between \(1000\) and \(2000\) which are *not* divisible by \(5\).

*are*divisible by \(5\) can be written as \[\begin{align*} 1000 + 1005 + \dotsb + 1995 + 2000 &= 5 \times 200 + 5 \times 201 + \dotsb + 5 \times 400 \\ &= 5 \times \left( 200 + 201 + \dotsb + 400 \right) \\ &= 5 \sum_{i=200}^{400} i \\ &= \frac{5 \times (200+400) \times (400 - 200 + 1)}{2} \\ &= \frac{5 \times 600 \times 201}{2} \\ &= 301\,500. \end{align*}\] Hence, the desired sum is \[\begin{equation*} 1\,501\,500 - 301\,500 = 1\,200\,000. \end{equation*}\]

- A geometric series has first term \(2\) and common ratio \(0.95\). The sum of the first \(n\) terms of the series is denoted by \(S_n\) and the sum to infinity is denoted by \(S\). Calculate the least value of \(n\) for which \(S - S_n < 1\).

We are to find the least value of \(n\) for which \(S - S_n < 1\); let’s try to solve \(S - S_n = 1\).

This gives us \(40 - 2\dfrac{1 - 0.95^n}{1-0.95}=40 - 40(1-0.95^n)=1\), and multiplying out and rearranging gives \[0.95^n=\dfrac{1}{40} \implies n = \dfrac{\ln(1/40)}{\ln 0.95} = 71.9...\]

Thus the value of \(n\) that we require is \(72\).

Alternatively, we could solve \(0.95^n=\dfrac{1}{40}\) by taking logs to base \(0.95\).