User:SahilkrM/sandbox

(Q)1.Using the method of mathematical induction,show that for all n € N, 1+2+3+...+n=n(n+1)/2

Soln: Let us take , S(n)=>1+2+3+...+n=n(n+1)/2,n € N

When,n=1,then, L.H.S.=n,  R.H.S.=n(n+1)/2 =1,        =1(1+1)/2                    =1×2/2                   =2/2                   =1

.°. L.H.S=R.H.S.,so,the statement given is true for n=1

Let,the statement be true for n=k, i.e S(k)=>1+2+3+...+k=k(k+1)/2(i)

Again,Let,the statement be true for n=k+1,i.e.

=>1+2+3+...+k+(k+1)=(k+1)(k+2)/2(ii)

Now,adding k+1 to both the sides of eqn.(i)

=>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+2)/2(iii)

.°.(iii)=(i)

Hence,the statement given for n=k+1 is also true.

So,it is true for all n € N                     #solved