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1+2+3+4+....+999 =
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1000*1000/2 = 500000
wait dammit 1000*999/2 = 499500
999*1000/2
1+2+3+4+...........+N =N(N+1)/2
\[\sum _{n=1}^{999} n=499500 \]
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1+2+3+. . . +n n+. . . +3+2+1 (n(n+1)) / 2
@kdrkitty Did you understand it? :)
1+2+3+...+999 999+998+...+2+1 and now add the elements vertically. This will yield 1000+1000+1000+.....+1000, which is just 999*1000 = 999000 Now this is twice the answer we wanted, so just divide by 2. The answer is 499500
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