数学有公式
9228+(9228-10)+(9228-20)+(9228-30)+(9228-40)+(9228-50)+(9228-60)+(9228-70)+(9228-80)+(9228-90)+(9228-100)+(9228-110)+(9228-120)=有公式吗?
(9228-0)+(9228-10)+(9228-20)+……+(9228-120)
=9228×13+(0+120)×13÷2
=119964+780
=120744
等差数列
a(1)=9228,d=-10,n=13
前n项和
S(n)=a(1)*n+dn(n-1)/2
根据拆括号解。
9228x13-130x5-60
9228+(9228-10)+(9228-20)+(9228-30)+(9228-40)+(9228-50)+(9228-60)+(9228-70)+(9228-80)+(9228-90)+(9228-100)+(9228-110)+(9228-120)
=9228×13-10×(1+2+3+4+.........+11+12)
=119964-10×12×(12+1)/2
=119964-780
=119184
等差数列求和公式
祝你好好学习,天天向上
原式=9228*13-10*{(1+12)*1/2}
用等差数列求和公式:Sn=(a1+an)*n/2
=9228*13-(10+20+30+...+120)
=119964-(10+120)*12/2
=119964-780
=119184
等差数列求和公式:(首项+末项)*项数/2
9228+(9228-10)+(9228-20)+(9228-30)+(9228-40)+(9228-50)+(9228-60)+(9228-70)+(9228-80)+(9228-90)+(9228-100)+(9228-110)+(9228-120)
=9228×13-10×(1+2+3+4+.........+11+12)
=119964-10×12×(12+1)/2
=119964-780
=119184
13×9288+(-10-120)×12/2
=9228*13-(10+20+30+40+50+60+70+80+90+100+110+120)
=9228*13-(10+120)*12/2
=9228*13-130*6
=13*(9228-60)
=13*9168
=119184
等差数列求和公式啊!就和1+2+3+……+100一样的。