The positive integer \(k\) is given.
- Find, in terms of \(k\), an expression for \(S_1\), the sum of the integers from \(2k\) to \(4k\) inclusive.
Approach 1
We could add together the integers \(2k\), \(2k+1\), \(\dotsc\), \(4k\) excluding \(3k\) in pairs as follows: \[\begin{align*} 2k+4k &= 6k \\ (2k+1)+(4k-1) &= 6k \\ \vdots \quad & \\ (3k-1)+(3k+1) &= 6k. \end{align*}\]And there are \(k\) of these pairs in total.
So to add the numbers between \(2k\) and \(4k\) inclusive, we can add \(k\) pairs that add to \(6k\), together with \(3k\), that is, \[S_1=6k^2+3k.\]
Approach 2
We could use the fact that the sum of the integers from \(1\) to \(n\) is \(P_n=\dfrac{n(n+1)}{2}\).
So we want to find \(P_{4k}-P_{2k-1}\).
- Find, in terms of \(k\), an expression for \(S_2\), the sum of the odd integers lying between \(2k\) and \(4k\).
The idea behind the first approach above can be used again for (a)(ii).
Approach 1
If \(k\) is even, we can add the odd integers between \(2k\) and \(4k\) in \(k/2\) pairs that add to \(6k\): \[\begin{align*} (2k+1)+(4k-1) &= 6k \\ (2k+3)+(4k-3) &= 6k \\ \vdots \quad & \\ (3k-1)+(3k+1) &= 6k. \end{align*}\] So \[\begin{align*} S_2 &= 6k\times\frac{k}{2}\\ &=3k^2. \end{align*}\]We can use exactly the same approach if \(k\) is odd, only this time the odd integers between \(2k\) and \(4k\) consist of \((k-1)/2\) pairs that add to \(6k\), together with \(3k=6k/2.\)
So once again \[S_2=3k^2.\]
Approach 2
We need \((2k+1) + (2k+3) + \cdots + (4k-1)\), or \((2k+1) + (2k+3) + \cdots + (2k+(2k-1))\).
This is an arithmetic sequence with \(k\) terms, so for the formula, \(n = k\), \(d = 2\) and \(a = 2k+1\).
The sum is therefore \(\dfrac{n}{2}(2a+(n-1)d) = \dfrac{k}{2}(4k+2+2(k-1)) = k(3k ) =3k^2\).
- Show that \(\dfrac{S_1}{S_2} = 2+\dfrac{1}{k}\).
- Prove that the sum of the first \(n\) terms of the geometric progression having first term \(a\) and common ratio \(r\) (\(r \neq 1\)) is \[a \left( \frac{1-r^n}{1-r} \right).\]
The sum of the first \(n\) terms of a geometric progression is \[T_n=a+ar+ar^2+\cdots+ar^{n-1}.\]
Now notice that \[r\times T_n = ar+ar^2+\cdots+ar^{n-1}+ar^n.\]
Subtracting, most of the terms cancel out so that \[\begin{align*} (1-r)T_n &= a-ar^n\\ \implies\quad T_n &= \frac{a(1-r^n)}{(1-r)}. \end{align*}\]By regarding the recurring decimal \(0.\dot{0}7\dot{5}\) (\(=0.075075\ldots\), where the figures \(075\) repeat) as an infinite geometric progression, or otherwise, obtain the value of the decimal as a fraction in its lowest terms.
Now \(0.075075\ldots\) can be written as the sum of a geometric progression with first term \(0.075\) and common ratio \(0.001\).
Using the expression above for the sum of the first \(n\) terms, and taking the limit as \(n\to \infty\), we have \[T_n\to a \left( \frac{1}{1-r} \right)\] (since \(|r|<1\), and \(r^n \to 0\) as \(n\to\infty\)).
Substituting in our values of \(a\) and \(r\) gives \[0.\dot{0}7\dot{5}=\frac{0.075}{1-0.001}=\frac{0.075}{0.999}=\frac{25}{333}.\]
Alternatively, let \(x = 0.075075\ldots\), so \(1000x = 75.075075\ldots\).
Subtracting, we get \(999x = 75\), so \(x=\dfrac{75}{999} = \dfrac{25}{333}\).
