Can we find the sum of the integers from $2k$ to $4k$ inclusive?

Can we find the sum of the integers from $2k$ to $4k$ inclusive?

Question

1. 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. Find, in terms of $k$, an expression for $S_2$, the sum of the odd integers lying between $2k$ and $4k$. Show that $\dfrac{S_1}{S_2} = 2+\dfrac{1}{k}$.

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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 $a \left( \frac{1-r^n}{1-r} \right).$ By regarding the recurring decimal $0.\dot{0}7\dot{5}$ ($=0.075075\dots c$, 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.

Suggestion

Is there a convenient way of dividing up these sums into smaller sums that are easier to deal with?

Can we use the fact that the $k$th term in this geometric progression has the form $a\times r^{k-1}$?

If we call the sum of the first n terms $T_n$, do you notice anything about the terms in $r\times T_n$?

What’s the first term of the geometric progression? What’s the common ratio?

Solution

Approach 1

We could add together the integers

$2k$

,

$2k+1$

,

$\dots c$

,

$4k$

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

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

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 $\dfrac{n}{2}(2a+(n-1)d) = \dfrac{k}{2}(4k+2+2(k-1)) = k(3k ) =3k^2$.

We have

$$\begin{align*} \frac{S_1}{S_2} &= \frac{6k^2+3k}{3k^2}\\ &=2+\frac{1}{k}. \end{align*}$$

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 $x=\dfrac{75}{999} = \dfrac{25}{333}$.