Question

What is the sum of the first \[n\] natural numbers \[1,2,3,4,...,n\]?

Answer 1

Hint Given question relates with the arithmetic progression. The terms of the given sequence are in arithmetic progression. By using the formula of sum of \[n\] terms in arithmetic progression, we will find the required answer.

Formula used
\[{n^{th}}\] term of an arithmetic progression: \[{a_n} = a + \left( {n - 1} \right)d\]
   Where \[a\] is the first term and \[d\] is the common difference
Sum of \[n\] terms in arithmetic progression: \[S = \dfrac{n}{2}\left( {\text{first term + last term}} \right)\]

Complete step by step solution:
The given sequence of \[n\] natural numbers is \[1,2,3,4,...,n\].
Clearly the sequence is an arithmetic progression.
Let \[a\] be the first term, \[d\] be the common difference and \[t\] be the last term of a finite arithmetic progression.
Then, the \[{n^{th}}\] term is: \[t = a + \left( {n - 1} \right)d\]
The sum of the \[n\] terms in arithmetic progression is:
\[S = a + \left( {a + d} \right) + \left( {a + 2d} \right) + ..... + \left( {a + \left( {n - 2} \right)d} \right) + \left( {a + \left( {n - 1} \right)d} \right)\]
\[ \Rightarrow \]\[S = a + \left( {a + d} \right) + \left( {a + 2d} \right) + ..... + \left( {t - d} \right) + t\] \[.....\left( 1 \right)\]
We know that addition is commutative.
Therefore,
\[S = t + \left( {t - d} \right) + ..... + \left( {a + 2d} \right) + \left( {a + d} \right) + a\] \[.....\left( 2 \right)\]
Now add the equation \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[S + S = \left[ {a + t} \right] + \left[ {\left( {a + d} \right) + \left( {t - d} \right)} \right] + \left[ {\left( {a + 2d} \right) + \left( {t - 2d} \right)} \right] + ..... + \left[ {\left( {t - d} \right) + \left( {a + d} \right)} \right] + \left[ {t + a} \right]\]
Simplify the above equation.
\[2S = \left( {a + t} \right) + \left( {a + t} \right) + \left( {a + t} \right) + ..... + \left( {a + t} \right) + \left( {a + t} \right)\]
\[ \Rightarrow \]\[2S = n\left( {a + t} \right)\]
\[ \Rightarrow \]\[S = \dfrac{{n\left( {a + t} \right)}}{2}\]
\[ \Rightarrow \]\[S = \dfrac{{n\left( {a + a + \left( {n - 1} \right)d} \right)}}{2}\] [ Since \[t = a + \left( {n - 1} \right)d\] ]
\[ \Rightarrow \]\[S = \dfrac{{n\left( {2a + \left( {n - 1} \right)d} \right)}}{2}\]
For first \[n\] natural numbers, the first term is \[a = 1\] and the common difference is \[d = 1\].
Substitute the values in the above equation.
\[S = \dfrac{{n\left( {2\left( 1 \right) + \left( {n - 1} \right)\left( 1 \right)} \right)}}{2}\]
Simplify the above equation.
\[S = \dfrac{{n\left( {2 + n - 1} \right)}}{2}\]
\[ \Rightarrow \]\[S = \dfrac{{n\left( {n + 1} \right)}}{2}\]
Hence, the sum of the first \[n\] natural numbers \[1,2,3,4,...,n\] is \[\dfrac{{n\left( {n + 1} \right)}}{2}\].

Note: Arithmetic progression is a sequence of numbers in which the difference between any two consecutive numbers is a constant value. The constant value is called a common difference.
The general form of an AP is \[a,\;a + d,\;a + 2d,\;a + 3d,.....\;\] where \[a\] is the first term, and \[d\] is a common difference.