1 + 2 + 3 + … + n = $$\dfrac{ n (n +1)}{2}$$ i.e., the sum of the first n natural numbers is $$\dfrac{ n (n + 1)}{2}$$

Asked by Aaryan | 1 year ago |  98

##### Solution :-

For, n = 1

LHS of P (n) = 1

RHS of P (n) = $$\dfrac{ 1 (1+1)}{2}$$ = 1

So, LHS = RHS

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

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

1 + 2 + 3 + …. + k = $$\dfrac{ k (k+1)}{2}$$ … (i)

Now,

(1 + 2 + 3 + … + k) + (k + 1)

$$\dfrac{ k (k+1)}{2}$$ + (k+1)

= (k + 1) ($$\dfrac{k}{2}$$ + 1)

=$$\dfrac{ [(k+1) [(k+1) + 1]] }{ 2}$$

P (n) is true for n = k + 1

P (n) is true for all n ∈ N

So, by the principle of Mathematical Induction

Hence, P (n) = 1 + 2 + 3 + ….. + n = $$\dfrac{ n (n +1)}{2}$$ is true for all n ∈ N.

Answered by Aaryan | 1 year ago

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