"Adding Consecutive Numbers"@en . "The count is easy and try but there is a faster shortcut. You can count 1+2+3+4+5+6+... tiring right. There a shortcut for it is to take as 1+2+3+4+...+n First take n/2 next take n+1 then mutiply (n/2)(n+1) is 1+2+3+...+n n is an algebrabic unit. try 10 10/2=5 10+1=11 5*11=55 or take 1+2+...+n-1+n = take half of n then add half then take yor number after that multipy (n/2+1/2)n if x = sum (1...n) then x + (n + 1) = n * (n+1) / 2 + (n + 1) x + (n + 1) = (n * (n + 1) + 2n +2) / 2 x + (n + 1) = ((n*n + n) + 2n + 2) / 2 x + (n + 1) = (n*n + 3 *n + 2) / 2 x + (n + 1) = ((n + 1) * (n + 2)) / 2 What if it doesn't start at 1? Like 15+16+17+... Minus the smaller consecutive numbers from the bigger ones. Try from 21 to 80. (80+1) * (80/2) - (20/2) * (20+1) =(81 * 40) - (10 * 21) =3240 - 210 =3030"@en . "The count is easy and try but there is a faster shortcut. You can count 1+2+3+4+5+6+... tiring right. There a shortcut for it is to take as 1+2+3+4+...+n First take n/2 next take n+1 then mutiply (n/2)(n+1) is 1+2+3+...+n n is an algebrabic unit. try 10 10/2=5 10+1=11 5*11=55 or take 1+2+...+n-1+n = take half of n then add half then take yor number after that multipy (n/2+1/2)n if x = sum (1...n) then x + (n + 1) = n * (n+1) / 2 + (n + 1) x + (n + 1) = (n * (n + 1) + 2n +2) / 2 x + (n + 1) = ((n*n + n) + 2n + 2) / 2 x + (n + 1) = (n*n + 3 *n + 2) / 2 x + (n + 1) = ((n + 1) * (n + 2)) / 2 Like 15+16+17+... =3030"@en . .