Question

How do you find the sum of the first $1000$ positive even integer?

Answer 1

Hint: In this question, we have been asked to find the sum of first $1000$ positive even integers. Since each term has the difference of $2$, they will form an A.P. Hence, we need to apply the sum of A.P., (arithmetic progression) formula to obtain the sum of the first $1000$ positive integers. A sequence of the following form
$a,a + d,a + 2d,a + 3d, \ldots ,a + (n - 1)d,a + nd, \ldots $
is called an arithmetic progression or an arithmetic sequence. In other words, each term (other than the first term) of the sequence is obtained by adding a constant to its previous term. The constant $d$ is called common difference and the term $a$ is called the initial term.
And then we use the formula for the sum $S$ of $n$ terms in A.P.,
$S = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$

Complete step by step answer:
The positive even integer n can be written as $n = 2 \cdot k$ , where $k$ is also an integer
The sum you are looking from is
$2 + 4 + 6 + 8 + \ldots + 2000$ ,
Now we will take $2$ as a common factor,
This is the same as
$2(1 + 2 + 3 + 4 + \ldots + 1000)$ …. (1)
The total number of terms in the series are $1000$
$n = 1000$
First number $a$ of the series is
$a = 1$
The common difference $d$ is
$d = 1$
The formula for the sum $S$of $n$terms in A.P.,
$S = \dfrac{n}{2}\left\{ {2a + (n - 1)d} \right\}$
Now, we substitute for $n$ , $a$ and $d$ in the formula for the sum of A.P .
The equation is,
$S = \dfrac{{1000}}{2}\left\{ {2(1) + (1000 - 1)1} \right\}$
We divide $1000$ by $2$ and subtract the value
$S = 500\left\{ {2 + 999} \right\}$
Now we add the set bracket
$S = 500\left\{ {1001} \right\}$
Putting this is equation (1)
\[2(500 \times 1001)\]
$1001000$ is a sum of number series by applying the values of input parameters in the formula.

Note: For $n \in \mathbb{N},{T_n} = an + b$ where $a$ and $b$ are relatively prime, form an AP which contains infinitely many prime numbers along with infinitely many composite numbers.
The ${n^{th}}$ term of an arithmetic progression is given by ${T_n} = a + (n - 1)d$
If you don’t know what will be the 1000th even number, you can use this formula to find the answer by keeping $a = 2$ , $n = 1000$ and $d = 2$ .You will find the 1000 even number.