Question

Calculate the sum of natural numbers between 1 and 42.
A) 460
B) 900
C) 903
D) 920

Answer 1

Hint:
We are given the two limits as 1 and 42. But here we are not going to literally add the numbers rather we will use the formula for finding the sum of n natural numbers.
\[sum = \dfrac{{n\left( {n + 1} \right)}}{2}\] where n is the numbers to be added. Let’s solve it!

Complete step by step solution:
Given that there are 42 numbers in 1 and 42. So n=42.
Now using the formula mentioned above
\[sum = \dfrac{{n\left( {n + 1} \right)}}{2}\]
\[ \Rightarrow \dfrac{{42\left( {42 + 1} \right)}}{2}\]
Divide 42 by 2 and add the numbers in bracket
\[ \Rightarrow 21 \times 43\]
On multiplying we get
\[ \Rightarrow 903\]
Hence 903 is the sum of natural numbers between 1 and 42.

So the correct option is C.

Note:
Alternate method:
Given numbers are in an arithmetic series like a sequence we can say. Where first term a=1 and common difference d=1. Isn’t it??
So e can use the formula of sum of terms of an arithmetic series as
\[{S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\]
Here a=1, d=1. So, it becomes
\[{S_n} = \dfrac{n}{2}\left( {2 + \left( {n - 1} \right)} \right)\]
\[ \Rightarrow {S_n} = \dfrac{n}{2}\left( {n + 1} \right)\]
And this is the same formula we used above.
\[ \Rightarrow {S_n} = \dfrac{{42}}{2}\left( {42 + 1} \right) = 21 \times 43 = 903\]