Prove the following by the principle of mathematical induction:1 + 2 +

1.	Prove the following by the principle of mathematical induction:1 + 2 + 3 +.....+ n = \(\frac{n(n+1)}2\) i.e., the sum of the first n natural numbers is \(\frac{n(n+1)}2\).
Answer» Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n = \(\frac{n(n+1)}2\) For n = 1 L.H.S of P(n) = 1 R.H.S of P(n) = \(\frac{1(1+1)}2\) = \(\frac{2}2\) = 1 Therefore, L.H.S = R.H.S Since, P(n) is true for n = 1 Let assume P(n) be the true for n = k, so 1 + 2 + 3 + - - - - - + k = \(\frac{k(k+1)}2\) - - - (1) Now (1 + 2 + 3 + - - + k) + (k + 1) = \(\frac{k(k+1)}2\) + (k + 1) = (k + 1)\((\frac{k}2+1)\) = \(\frac{(k+1)(k+2)}2\) = \(\frac{(k+1)[(k+1)+1]}2\) 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 = \(\frac{n(n+1)}2\) is true for all n∈N

Prove the following by the principle of mathematical induction:1 + 2 + 3 +.....+ n = \(\frac{n(n+1)}2\) i.e., the sum of the first n natural numbers is \(\frac{n(n+1)}2\).

Answer»

Let us Assume P(n) = 1 + 2 + 3 + - - - - - - + n = \(\frac{n(n+1)}2\)

For n = 1

L.H.S of P(n) = 1

R.H.S of P(n) = \(\frac{1(1+1)}2\) = \(\frac{2}2\) = 1

Therefore, L.H.S = R.H.S

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

Let assume P(n) be the true for n = k,

1 + 2 + 3 + - - - - - + k = \(\frac{k(k+1)}2\) - - - (1)

Now

(1 + 2 + 3 + - - + k) + (k + 1)

= \(\frac{k(k+1)}2\) + (k + 1)

= (k + 1)\((\frac{k}2+1)\)

= \(\frac{(k+1)(k+2)}2\)

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

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 = \(\frac{n(n+1)}2\) is true for all n∈N