Question

What is the sum of the series: $1 + 2 + 3 + 4 + 5 + ..... + n$ ?

Answer 1

Hint:The given problem requires us to find the sum of an arithmetic progression. The first few terms and the last term of the series is given to us in the question. For finding out the sum of an arithmetic progression, we need to know the first term, the common difference and the number of terms in the arithmetic progression. We can find out the common difference of an arithmetic progression by knowing the difference of any two consecutive terms of the series.

Complete step by step answer:
So, we have, $1 + 2 + 3 + 4 + 5 + ..... + n$
The difference of any two consecutive terms of the given series is constant. So, the given sequence is an arithmetic progression.
Now, we have to find the sum of this arithmetic progression.
Here, first term $ = a = 1$.
Now, we can find the common difference of the arithmetic progression by subtracting any two consecutive terms in the series.
So, common ratio \[ = d = 2 - 1 = 1\]
So, $d = 1$ .
We also need to know the number of terms in the arithmetic progression in order to find the value of the sum of arithmetic progression.

Since the first term is a and the last term is $n$. Also, the common difference between the terms is one. Hence, the number of terms is $n$. Now, we can find the sum of the given arithmetic progression using the formula $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Hence, the sum of AP $ = S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
Substituting the values of a, d and n in the formula, we get,
$ \Rightarrow S = \dfrac{n}{2}\left[ {2\left( 1 \right) + \left( {n - 1} \right)\left( 1 \right)} \right]$
Opening the brackets simplifying the expression, we get,
$ \Rightarrow S = \dfrac{n}{2}\left[ {2 + n - 1} \right]$
Simplifying the expression further, we get,
$ \therefore S = \dfrac{{n\left( {n + 1} \right)}}{2}$

So, the sum of the given series: $1 + 2 + 3 + 4 + 5 + ..... + n$ is $S = \dfrac{{n\left( {n + 1} \right)}}{2}$.

Note:Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The sum of n terms of an arithmetic progression can be calculated if we know the first term, the number of terms and difference of the arithmetic series as: $S = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.