Question

$class \: 11$

Answer 1

Hi there,

The Natural No.s series is given by
1, 2, 3, 4, ………n terms

Here the series is in A.P
a= 1
d= 1

Sum of terms in an A.P= (n/2)×[2a+(n-1)d]

•°• Sum of first n Natural Numbers
= (n/2)×[(2×1)+((n-1)×1)]
= (n/2)×[2+n-1]
= (n/2)×(n-1)
= [n(n-1)]/2



;)
hope it helps

Answer 2

Let  P(n):-$1+2+3+4+............n=\frac{n(n+1)}{2}$

Step:-1

We have,

P(1): 1= {1(1+1)}/2

      1 = 1

So, P(1) is true

Step:-2

Let P(m) be true;

$1+2+3+.......................+m=\frac{m(m+1)}{2}\;\;\;\;\;..............(i)$

Step:-3

We have  to show P(m+1) is true,For  this  we have to show that,

1+2+3+4............m+m+1=$\frac{(m+1)[(m+1)+1]}{2}$

Now,

$1+2+3+4+...........m+m+1\\\\=\frac{m(m+1)}{2}+m+1\\\\=\frac{m(m+1)+2(m+1)}{2}\\\\=\frac{(m+1)(m+2)}{2}\\\;\\=\frac{(m+1)[(m+1)+1]}{2}$

Thus,

P(m) is true ⇒ P(m+1) is true

Hence,By  the Principle of Induction, the given result  is  true  for all natural numbers.