Question

Find the sum of first n natural numbers.

Answer 1

Hint: Start by identifying and writing the first n natural numbers and look for the sequence formed I.e. Arithmetic progression and distinguish the first term , common difference and total number of terms , apply the general formula for sum of first n terms of an A.P. and simplify for the final answer.

Complete step-by-step answer:
In order to find the sum of first n natural numbers first of all we have to find the first n natural numbers.
And we know the first natural numbers are 1, 2, 3, 4, 5……. (n-1), n
So now clearly if you see they are in AP (arithmetic progression)
Whose first term is a = 1 and
common difference is d = 1
Let ${S_n}$ denote the sum of first n natural numbers
i.e. \[{S_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]\]
Where a is the first term and n is the number of terms and d is the common difference.
Substituting all the values of a , d , n , we get
$
  {S_n} = \dfrac{n}{2}\left[ {2 \times 1 + (n - 1) \times 1} \right] \\
  {S_n} = \dfrac{n}{2}\left[ {2 + n - 1} \right] \\
  \therefore {S_n} = \dfrac{n}{2}\left[ {n + 1} \right] \\
$
Hence, The sum of first n natural numbers will be \[{S_n} = \dfrac{n}{2}\left[ {n + 1} \right]\]

Note: The above derived answer is an important formula and must be remembered which makes the calculation easier. Such similar questions can be solved by using the same procedure as above. Students must be well aware of all the terminologies used in different sequences and series. Questions can also be asked as sum of first n even natural numbers, In that case a = 2 , d = 2 .