Review question

# Can we find the sum of the integers from $2k$ to $4k$ inclusive? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5849

## Solution

1. The positive integer $k$ is given.

1. 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 $P_{4k}-P_{2k-1}$.

We have \begin{align*} S_1 &= P_{4k}-P_{2k-1} \\ &=\frac{1}{2} (4k)(4k+1) - \frac{1}{2} (2k-1)(2k)\\ &=6k^2+3k. \end{align*}
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 $\dfrac{n}{2}(2a+(n-1)d) = \dfrac{k}{2}(4k+2+2(k-1)) = k(3k ) =3k^2$.

1. Show that $\dfrac{S_1}{S_2} = 2+\dfrac{1}{k}$.
We have \begin{align*} \frac{S_1}{S_2} &= \frac{6k^2+3k}{3k^2}\\ &=2+\frac{1}{k}. \end{align*}
1. 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 $x=\dfrac{75}{999} = \dfrac{25}{333}$.