Question

How do you find the sum of the first $5000$ natural numbers?

Answer 1

Natural numbers are counting numbers starting from 1. First we have to identify the type of series that will be formed according to the question. Then we can use the formula to find the sum of the series up to $n$ terms. We do not have to find all the terms in order to find the sum of the series.

Formula used:
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$

Complete step by step solution:
The natural numbers can be written in this manner $= 1,2,3,4,5,6...$
For the first $5000$ natural numbers, we can write the above series as $= 1,2,3,4,5,6,...,5000$
First we have to identify the type of the given series, i.e. whether the series is Arithmetic Progression or Geometric Progression.
We can observe that the difference between the consecutive terms is common for all the terms in the series, which is $2 - 1 = 3 - 2 = 4 - 3 = 1$.
Thus, the given series is an Arithmetic Progression (AP) with common difference $d = 1$.
Now we can use the formula to find the sum of the series up to $5000$ terms. The formula of the sum of the series up to $n$ terms is given by,
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
where, ${S_n}$ is the sum of the series up to $n$ terms
$a$ is the first term of the series
$d$ is the common difference
$n$ is the number of the terms up to which sum is to be calculated.
In the given series, $a = 1$ and $d = 1$.
We have to find sum up to $5000$ terms, so $n = 5000$.
Putting all the values in the above formula, we get:

$${S_{5000}} = \dfrac{{5000}}{2}[2 \times 1 + (5000 - 1) \times 1] \\\ \Rightarrow {S_{5000}} = 2500 \times [2 + 4999 \times 1] \\\ \Rightarrow {S_{5000}} = 2500 \times [2 + 4999] \\\ \Rightarrow {S_{5000}} = 2500 \times 5001 \\\ \Rightarrow {S_{5000}} = 12502500 \\\$$

Thus, the sum of the given series up to $5000$ terms is $12502500$.

Note: We can find the sum of the series up to $n$ terms using the formula without knowing all the terms. We only need at most three terms to identify the type of series and calculate the common difference, using which we can calculate the sum of the series up to $n$ terms, where $n \geqslant 1$.