七年级数学有理数加减法简便计算练习题及参考答案C6

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+···+2208。(2)(-1)+(-2)+(-3)+(-4)+···+(-524)。(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+1165)+(+1166)+(-1167)+(-1168)。(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-2961)+(-2962)+(+2963)+(+2964)。(5)(-1)+(+2)+(-3)+(+4)+···+(-2077)+(+2078)。(6)(-2006)+(+2005)+···+(-4)+(+3)+(-2)+(+1)。(7)1+(-2)+3+(-4)+···+2317+(-2318)。(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,4154×4155)(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,5177×5179)(10)eq\f(1,6)+eq\f(1,36)+eq\f(1,216)+eq\f(1,1296)+eq\f(1,7776)。二、详细步骤(1)1+2+3+4+···+2208。解：原式=(1+2208)+(2+2207)+(3+2206)+···+(1104+1105)=(1+2208)*1104=2438736.(2)(-1)+(-2)+(-3)+(-4)+···+(-524)。解：原式=-(1+2+3+4+···+524)=-[(1+524)+(2+523)+(3+522)+···+(262+263)]=-[(1+524)*262]=-137550.(3)(+1)+(+2)+(-3)+(-4)+(+5)+(+6)+(-7)+(-8)+···+(+1165)+(+1166)+(-1167)+(-1168)。解：原式=[(+1)+(+2)+(-3)+(-4)]+[(+5)+(+6)+(-7)+(-8)]+···+[(+1165)+(+1166)+(-1167)+(-1168)]=(1+2-3-4)+(5+6-7-8)+···+(1165+1166-1167-1168)=-4+(-4)+···+(-4)=-4*292=-1168.(4)(-1)+(-2)+(+3)+(+4)+(-5)+(-6)+(+7)+(+8)+···+(-2961)+(-2962)+(+2963)+(+2964)。解：原式=[(-1)+(-2)+(+3)+(+4)]+[(-5)+(-6)+(+7)+(+8)]+···+[(-2961)+(-2962)+(+2963)+(+2964)]=(-1-2+3+4)+(-5-6+7+8)+···+(-2961-2962+2963+2964)=4+4+···+4=4*741=2964.(5)(-1)+(+2)+(-3)+(+4)+···+(-2077)+(+2078)。解：原式=[(-1)+(+2)]+[(-3)+(+4)]···+[(-2077)+(+2078)]=(-1+2)+(-3+4)···+(-2077+2078)=1+1+···+1=1*1039=1039.(6)(-2006)+(+2005)+···+(-4)+(+3)+(-2)+(+1)。解：原式=[(-2006)+(+2005)]+···+[(-4)+(+3)]+[(-2)+(+1)]=(-2006+2005)+···+(-4+3)+(-2+1)=-1+(-1)+···+(-1）=-1*1003=-1003.(7)1+(-2)+3+(-4)+···+2317+(-2318)。解：原式=[1+(-2)]+[3+(-4)]···+[2317+(-2318)]=(1-2)+(3-4)···+(2317-2318)=-1+(-1)+···+(-1)=-1*1159=-1159.(8)eq\f(1,1×2)+eq\f(1,2×3)+eq\f(1,3×4)+···+eq\f(1,4154×4155)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(1,4154)-eq\f(1,4155))=1-eq\f(1,2)+eq\f(1,2)-eq\f(1,3)+eq\f(1,3)-eq\f(1,4)+···+eq\f(1,4154)-eq\f(1,4155)=1-eq\f(1,4155)=eq\f(4154,4155).(9)eq\f(1,1×3)+eq\f(1,3×5)+eq\f(1,5×7)+···+eq\f(1,5177×5179)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(1,5177)-eq\f(1,5179))]=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(1,5177)-eq\f(1,5179))=eq\f(1,2)·(1-eq\f(1,5179))=eq\f(2589,5179).(10)eq\f(1,6)+eq\f(1,36)+eq\f(1,216)+eq\f(1,1296)+eq\f(1,7776)。解：原式=G=eq\f(1,6)+eq\f(1,36)+eq\f(1,216)+eq\f(1,1296)+eq\f(1,7776),两边同