Verified Answer :

caylus

Hello,

\(s=1+2+3+...+1000+1001=\sum_{i=1}^{1001}\ i \ (1) \\ s=1001+1000+...+3+2+1=\sum_{j=1}^{1001}\ 1002-j \ (2)\\ (1)+(2)==\textgreater \ 2*s=(1+1001)+(2+1000)+...+(1000+2)+(1001+1)\\ =1002*1001=(\sum_{i=1}^{1001}\ i )\ +(\sum_{i=1}^{1001}\ 1002-i )\ \\ =\sum_{i=1}^{1001}\ ( i+1002-i)\ = \sum_{i=1}^{1001}\ 1002=1001*1002\\ So\ s= \dfrac{1001*1002}{2} =501501\\\)

Related Questions