七年级数学有理数加减法简便计算练习题及参考答案C4

七年级数学有理数加减法简便计算练习题及答案简要步骤姓名：班级：一、计算练习题(1)1+2+3+4+···+1864。(2)(-1)+(-2)+(-3)+(-4)+···+(-28)。(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+2429)+(+2430)+(-2431)+(-2432)。(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-1957)+(-1958)+(+1959)+(+1960)。(5)(-1)+(+2)+(-3)+(+4)+···+(-1009)+(+1010)。(6)(-24)+(+23)+···+(-4)+(+3)+(-2)+(+1)。(7)1+(-2)+3+(-4)+···+77+(-78)。(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,2794×2795)(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,3031×3033)(10)eq\f(1,3)+eq\f(1,9)+eq\f(1,27)+eq\f(1,81)+eq\f(1,243)。二、详细步骤(1)1+2+3+4+···+1864。解：原式=(1+1864)+(2+1863)+(3+1862)+···+(932+933)=(1+1864)*932=1738180.(2)(-1)+(-2)+(-3)+(-4)+···+(-28)。解：原式=-(1+2+3+4+···+28)=-[(1+28)+(2+27)+(3+26)+···+(14+15)]=-[(1+28)*14]=-406.(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+2429)+(+2430)+(-2431)+(-2432)。解：原式=[(+1)+(+2)+(-3)+(-4)]+[(+5)+(+6)+(-7)+(-8)]+···+[(+2429)+(+2430)+(-2431)+(-2432)]=(1+2-3-4)+(5+6-7-8)+···+(2429+2430-2431-2432)=-4+(-4)+···+(-4)=-4*608=-2432.(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-1957)+(-1958)+(+1959)+(+1960)。解：原式=[(-1)+(-2)+(+3)+(+4)]+[(-5)+(-6)+(+7)+(+8)]+···+[(-1957)+(-1958)+(+1959)+(+1960)]=(-1-2+3+4)+(-5-6+7+8)+···+(-1957-1958+1959+1960)=4+4+···+4=4*490=1960.(5)(-1)+(+2)+(-3)+(+4)+···+(-1009)+(+1010)。解：原式=[(-1)+(+2)]+[(-3)+(+4)]···+[(-1009)+(+1010)]=(-1+2)+(-3+4)···+(-1009+1010)=1+1+···+1=1*505=505.(6)(-24)+(+23)+···+(-4)+(+3)+(-2)+(+1)。解：原式=[(-24)+(+23)]+···+[(-4)+(+3)]+[(-2)+(+1)]=(-24+23)+···+(-4+3)+(-2+1)=-1+(-1)+···+(-1）=-1*12=-12.(7)1+(-2)+3+(-4)+···+77+(-78)。解：原式=[1+(-2)]+[3+(-4)]···+[77+(-78)]=(1-2)+(3-4)···+(77-78)=-1+(-1)+···+(-1)=-1*39=-39.(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,2794×2795)解：原式=1-eq\f(1,2)+(eq\f(1,2)-eq\f(1,3))+(eq\f(1,3)-eq\f(1,4))+···+(eq\f(1,2794)-eq\f(1,2795))=1-eq\f(1,2)+eq\f(1,2)-eq\f(1,3)+eq\f(1,3)-eq\f(1,4)+···+eq\f(1,2794)-eq\f(1,2795)=1-eq\f(1,2795)=eq\f(2794,2795).(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,3031×3033)解：原式=eq\f(1,2)·[(1-eq\f(1,3))+(eq\f(1,3)-eq\f(1,5))+(eq\f(1,5)-eq\f(1,7))+···+(eq\f(1,3031)-eq\f(1,3033))]=eq\f(1,2)·(1-eq\f(1,3)+eq\f(1,3)-eq\f(1,5)+eq\f(1,5)-eq\f(1,7)+···+eq\f(1,3031)-eq\f(1,3033))=eq\f(1,2)·(1-eq\f(1,3033))=eq\f(1516,3033).(10)eq\f(1,3)+eq\f(1,9)+eq\f(1,27)+eq\f(1,81)+eq\f(1,243)。解：原式=W=eq\f(1,3)+eq\f(1,9)+eq\f(1,27)+eq\f(1,81)+eq\f(1,243)