1+2+3+...+n=(n(n+1))/2;
n=1, 1=(1(1+1))/2; 1=1.
n=k, 1+2+3+...+k=(k(k+1))/2.
n=k+1, 1+2+3+...+k+(k+1)=
((k+1)(k+1+1))/2;
((k+1)(k+1+1))/2=((k+1)(k+2))/2=
(k(k+1)+2(k+1))/2=(k(k+1))/2+(k+1).
[1+2+3+...+k]+(k+1)=(k(k+1))/2+(k+1),
1+2+3+...+k=((k(k+1))/2.
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Answers & Comments
1+2+3+...+n=(n(n+1))/2;
n=1, 1=(1(1+1))/2; 1=1.
n=k, 1+2+3+...+k=(k(k+1))/2.
n=k+1, 1+2+3+...+k+(k+1)=
((k+1)(k+1+1))/2;
((k+1)(k+1+1))/2=((k+1)(k+2))/2=
(k(k+1)+2(k+1))/2=(k(k+1))/2+(k+1).
[1+2+3+...+k]+(k+1)=(k(k+1))/2+(k+1),
1+2+3+...+k=((k(k+1))/2.