- Prove that the sum of all the integers between \(m\) and \(n\) inclusive (\(m, n \in \mathbb{Z}_+\), \(n > m\)) is \(\tfrac{1}{2}(m+n)(n-m+1)\).
We know that
\(1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}\). So we can write
\[\begin{align*}
m+(m+1) + (m+2) + \ldots + n &= (1 + 2 + \ldots + n)-(1 + 2 + \ldots + (m-1))\\
&= \frac{n(n+1)}{2} - \frac{(m-1)m}{2} \\
&= \frac{n^2 + n - m^2 + m}{2} \\
&= \frac{(m+n)(n-m+1)}{2}.
\end{align*}\]
Find the sum of all the integers between \(1000\) and \(2000\) which are not divisible by \(5\).
We can calculate this by summing all the integers between
\(1000\) and
\(2000\), and then subtracting the sum of all those divisible by
\(5\). By the above formula,
\[\begin{equation*}
\sum_{i=1000}^{2000} i = \frac{(1000 + 2000)(2000 - 1000 + 1)}{2} = \frac{3000 \times 1001}{2} = 1\,501\,500.
\end{equation*}\]
The sum of all numbers between
\(1000\) and
\(2000\) that
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 have
\[\begin{equation*}
S_n = a \frac{1 - r^n}{1-r},
\end{equation*}\]
where
\(a = 2\) and
\(r = 0.95\), and
\[\begin{equation*}
S = \lim_{n \to \infty} S_n = \frac{a}{1-r} = \dfrac{2}{0.05} = 40.
\end{equation*}\]
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 \(0.95\).
