Question

1+2-3-4+5+6-7-8+9+10…+2021的简便运算？

Answer 1

1+2+3+4+5+6+7+8+9+10+11+12+13=(1+13) ×〔 (13-1)/2+1/2〕=14 × (6+1/2)=14 × 6+14 × 1 / 2 = 84 + 7 = 91 the simple calculation method of this problem is that a maximum addend 13 plus a minimum addend 1 is multiplied by [(13-1) / 2 + 1 / 2] (because the 13 numbers are added together, and the last one remains, and this number must be 1 / 2 of (13 + 1)). This operation is simpler than 1 + 2 + 3 +... 13 direct operation

Answer 2

1+2+3+4+5+6+7+8+9+10+11+12+13=(1+13) ×〔 (13-1)/2+1/2〕=14 × (6+1/2)=14 × 6+14 × 1 / 2 = 84 + 7 = 91 Den enkle beregningsmetode for dette problem er, at en maksimal tillæg 13 plus en minimumtillæg 1 ganges med [(13-1) / 2 + 1 / 2] (fordi de 13 tal lægges sammen, og det sidste er tilbage, og dette tal skal være 1 / 2 af (13 + 1)). Denne operation er enklere end 1 + 2 + 3 +... 13 direkte drift

Answer 3

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 12 + 13 = (1 + 13) ×〔 (13 - 1) / 2 + 1 / 2 = 14 × (6 + 1 / 2) = 14 × 6 + 14 × 1 / 2 = 84 + 7 = 91 Простейшим методом вычисления является максимальное прибавление thirteen плюс минимальное прибавление 1, умноженное на [(13 - 1) / 2 + 1 / 2] ( из - за сложения thirteen цифр, последнее из которых должно составлять 1 / 2). Эта операция проста, чем прямая операция 1 + 2 + 3 +...13

Answer 4

公差为1的等差数列
若一个等差数列的首项为a1,末项an,Sn=n*(a1+an)/2.
1+2+3+4+……+2021=(1+2021)*2021/2=2043231

Answer 5

1+2+3+4+5+6+7+8+9+10+11+12+13=(1+13) ×〔 (13-1)/2+1/2〕=14 × (6+1/2)=14 × 6+14 × 1 / 2 = 84 + 7 = 91 il metodo di calcolo semplice di questo problema è che un addend massimo 13 più un addend minimo 1 è moltiplicato per [(13-1) / 2 + 1 / 2] (perché i 13 numeri sono sommati insieme, e l'ultimo rimane, e questo numero deve essere 1 / 2 di (13 + 1)). Questa operazione è più semplice di 1 + 2 + 3 +... 13 operazione diretta