Math, asked by raja0567, 5 months ago

Sum of first n natural numbers

Answers

Answered by MotiSani
2

Given:

First n natural numbers

To Find:

Sum of first n natural numbers

Solution:

In the number system, all the positive integers from 1 to infinity are termed as natural numbers.

Let Sn the sum of first n natural numbers

⇒ Sn = 1 + 2 +3 +.......+ n

Sn = \frac{n}{2}[2a + (n - 1)d] --------------------(a)

Putting a =1 and d = (2 - 1) =1 in eq (a)

Sn = \frac{n}{2} [2x1 + (n -1)1]

Sn = \frac{n}{2} (n + 1)

Hence the sum of first n natural numbers is  \frac{n}{2}(n + 1).

Answered by pulakmath007
4

SOLUTION

TO DETERMINE

Sum of first n natural numbers

CONCEPT TO BE IMPLEMENTED

If in an arithmetic progression

First term & Last term is given

Then the sum

  \displaystyle \sf{ =  \frac{n}{2} \bigg[First  \: term + Last \: term \bigg]  }

Where n is the number of terms in the progression

EVALUATION

Here we have to find the Sum of first n natural numbers

Then the progression is 1 , 2, 3 , ...... , n

First term = a = 1

Second term = 2

Third term = 3

∴ Second Term - First term = Third term - Second term

Hence it is an arithmetic progression

Number of terms = n

Hence the required sum

  \displaystyle \sf{ =  \frac{n}{2} \bigg[First  \: term + Last \: term \bigg]  }

  \displaystyle \sf{ =  \frac{n}{2} \bigg[1  + n \bigg]  }

  \displaystyle \sf{ =  \frac{n(n + 1)}{2}  }

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