Can we sum from $1000$ to $2000$ excluding multiples of 5?

Can we sum from $1000$ to $2000$ excluding multiples of 5?

Question

1. 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 $5$.

---

1. 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 $S - S_n < 1$.

Solution

We know that

$1 + 2 + \ldots + n = \dfrac{n(n+1)}{2}$

. So we can write

$$\begin{align*} m+(m+1) + (m+2) + \ldots + n &= (1 + 2 + \ldots + n)-(1 + 2 + \ldots + (m-1))\\ &= \frac{n(n+1)}{2} - \frac{(m-1)m}{2} \\ &= \frac{n^2 + n - m^2 + m}{2} \\ &= \frac{(m+n)(n-m+1)}{2}. \end{align*}$$

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

$$\begin{align*} 1000 + 1005 + \dots b + 1995 + 2000 &= 5 \times 200 + 5 \times 201 + \dots b + 5 \times 400 \\ &= 5 \times \left( 200 + 201 + \dots b + 400 \right) \\ &= 5 \sum_{i=200}^{400} i \\ &= \frac{5 \times (200+400) \times (400 - 200 + 1)}{2} \\ &= \frac{5 \times 600 \times 201}{2} \\ &= 301\,500. \end{align*}$$

Hence, the desired sum is

$$\begin{equation*} 1\,501\,500 - 301\,500 = 1\,200\,000. \end{equation*}$$

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 $0.95$.