Prove that: 1 + 2 + 3 + ... + n = (n(n+1))/2 i.e., the sum of the first n natural numbers is (n(n+1))/2.

Gracie Steele

Gracie Steele

Answered question

2023-01-05

Prove that: 1 + 2 + 3 + . . . . . . . . . + n = n ( n + 1 ) 2 i.e., the sum of the first n natural numbers is n ( n + 1 ) 2 .

Answer & Explanation

mercerizoblx

mercerizoblx

Beginner2023-01-06Added 7 answers

Let P(n) : 1 + 2 + 3 + . . . . . . . . . + n = n ( n + 1 ) 2

For n = 1,
LHS of P ( n ) = 1
RHS of P ( n ) = 1 ( 1 + 1 ) 2 1 = 1
Since, LHS = RHS
⇒ P(n) is true for n = 1
Let P(n) be true for n = k, so
1 + 2 + 3 + . . . . . . . . + k = k ( k + 1 ) 2 ......(1)
Now
( 1 + 2 + 3 + . . . . . . . + k ) + ( k + 1 )
= k ( k + 1 ) 2 + ( k + 1 )
= ( k + 1 ) ( k 2 + 1 )
= ( k + 1 ) ( k + 2 ) 2
= ( k + 1 ) [ ( k + 1 ) + 1 ] 2
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for all n ϵ N
SO, according to the mathematical induction principle
P ( n ) : 1 + 2 + 3 + . . . . . . . . + n = n ( n + 1 ) 2 is true for all n ϵ N.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?