Question

prove by mathematical induction that

Answer 1

Given,
1+2+3+....+n = $\frac{n(n+1)}{2}$

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

For n = 1,

LHS = 1, RHS = $\frac{n(n+1)}{2}$ =$\frac{1(1+1)}{2}$ = 1

Thus, LHS = RHS

Thus, the statement is true for n=1.

Let the statement be true for n=k.

Thus, S(k): 1+2+3+.....k = $\frac{k(k+1)}{2}$ .........(i)

To prove that the statement is true for n = k+1

Now,
We have to put k+1 in the place of last digit i.e. k and add it to both sides.

Thus,
(i) ⇒ 1+2+3...+k+(k+1) = $\frac{k(k+1)}{2}$ + (k+1)
= (k+1)[k/2 + 1]
= (k+1)(k+2)/2
= $\frac{(k+1)[(k+1)+1]}{2}$

Thus, the statement is true for n=k+1

Hence, the statement is true for any value of n, n ∈ N

Answer 2

✨Refer The Attachment✨