Prove by the principle of mathematical induction:1 + 2 + 3 + … + n =

1.	Prove by the principle of mathematical induction:1 + 2 + 3 + … + n = n(n +1)/2 i.e., the sum of the first n natural numbers is n(n + 1)/2.
Answer» Let us considering P(n) = 1 + 2 + 3 + ….. + n = n(n +1)/2 For the n = 1 LHS of P (n) = 1 RHS of P (n) = 1 (1+1)/2 = 1 Therefore, LHS = RHS Here, P (n) is true for n = 1 Let us consider P (n) be the true for n = k, therefore 1 + 2 + 3 + …. + k = k (k+1)/2 … (i) Then, (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 ∈ N Therefore, by the principle of Mathematical Induction Thus, P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2 is true for all n ∈ N.

Prove by the principle of mathematical induction:1 + 2 + 3 + … + n = n(n +1)/2 i.e., the sum of the first n natural numbers is n(n + 1)/2.

Answer»

Let us considering P(n) = 1 + 2 + 3 + ….. + n = n(n +1)/2

For the n = 1

LHS of P (n) = 1

RHS of P (n) = 1 (1+1)/2 = 1

Therefore, LHS = RHS

Here, P (n) is true for n = 1

Let us consider P (n) be the true for n = k, therefore

1 + 2 + 3 + …. + k = k (k+1)/2 … (i)

Then,

(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 ∈ N

Therefore, by the principle of Mathematical Induction

Thus, P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2 is true for all n ∈ N.