Question

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

Answer 1

Hint: In this question we are asked to find the sum of first 1000 even positive integers, we will make use of the sum of \[n\] terms in the arithmetic progression, which is given by the formula, \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\], where \[a\] is the first term, \[n\] is the total number of terms, and \[d\] is the common difference. So here we are asked first 1000 even integers and even integers start from 2, and continue as 2,4,6,8,……..,. So from this series we will find the first term, common difference and substitute the values in the formula to get the required result.

Complete step-by-step solution:
Given that we have to find the sum of first 1000 even positive integers,
We know that even integers start from 2, and continue as 2, 4, 6, 8… , the series is in arithmetic progression , where 2 is the first term , common difference will be \[4 - 2 = 2\], and number of terms will be equal to 1000,
Now using the sum of \[n\] terms in the arithmetic progression, which is given by the formula,\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\],
Here \[a = 2\], \[n = 1000\] and \[d = 2\],
By substituting the values in the formula we get,
$\Rightarrow$\[{S_{1000}} = \dfrac{{1000}}{2}\left[ {2(2) + \left( {1000 - 1} \right)2} \right]\],
Now simplifying we get,
$\Rightarrow$\[{S_{1000}} = 500\left[ {4 + \left( {999} \right)2} \right]\],
Again multiplying and removing the brackets, we get,
$\Rightarrow$\[{S_{1000}} = 500\left[ {4 + 1998} \right]\],
Now further simplifying we get,
$\Rightarrow$\[{S_{1000}} = 500\left[ {2002} \right]\],
Now removing the brackets by multiplying we get,
$\Rightarrow$\[{S_{1000}} = 1001000\],
So the sum of the first $1000$ even positive integers is $1001000$.

\[\therefore \] The sum of the first 1000 even positive integers is 1001000.

Note: An arithmetic progression is a sequence where the differences between every two consecutive terms are the same. An arithmetic progression is a sequence where each term, except the first term, is obtained by adding a fixed number to its previous term.
Here are some formulas related to arithmetic progression:
Sum of \[n\] terms=\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\],
\[{n^{th}}\] term is given by \[{a_n} = a + \left( {n - 1} \right)d\],
Common difference \[\left( d \right)\] is given by \[{a_{n + 1}} - {a_n}\].