Question

Find the sum of the first 100 positive integers.

Answer 1

Hint: The sum of n positive terms means we have to find the sum of the first 100 natural terms of the series. As we know that, the terms in natural series are in arithmetic form so we can find the sum by using the arithmetic sum formula which is: $ {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) $

Complete step-by-step answer:
Here, we have to take the first 100 positive integers.
So the series is: 1, 2, 3, 4… 98, 99, 100.
So here the first term of the series is 1.
The common difference between the series is: $ {\rm{second}}\;{\rm{term - first}}\;{\rm{term = 2 - 1 = 1}} $
The total number of terms taken in the series for the sum is 100.
Hence the sum of the 100 terms by using the formula $ {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right) $
 Where, n is the number of terms of series, a is the first term and d is the common difference of the series.
Now, substituting the values in the equation or formula then we get,
\[{S_n} = \dfrac{{100}}{2}\left( {2 \times 1 + \left( {100 - 1} \right) \times 1} \right)\]
Solving the equation by using the BODMAS method so first we open the brackets given in the equation,
 $
\Rightarrow {S_n} = \dfrac{{100}}{2}\left( {2 + \left( {99} \right) \times 1} \right)\\
\Rightarrow {S_n} = \dfrac{{100}}{2}\left( {2 + 99} \right)\\
\Rightarrow {S_n} = \dfrac{{100 \times 101}}{2}
 $
Dividing 100 by 2 then we get,
 $ {S_n} = 50 \times 101 $
Now, multiply 50 with 101 then we get,
 $ {S_n} = 5050 $
Therefore, the sum of 100 first positive integers is 5050.

Note: In this type of problem, when we have the series in arithmetic form and having common difference as 1 then the sum of its n terms can be carried out directly by using the formula $ \dfrac{{n(n + 1)}}{2} $ .
As in this problem we have the value of n as 100 then the sum of n term by using $ \dfrac{{n(n + 1)}}{2} $ formulas is $ \dfrac{{100(100 + 1)}}{2} = 50 \times 101 = 5050 $ .