急 计算49²-50²+51²-52²+……+97²-98²简算

问题描述:

急 计算49²-50²+51²-52²+……+97²-98²简算

49^2-50^2=(49-50)*(49+50)=-99,51²-52²==(51-52)*(51+52)=-103,为一等差数列

49²-50²+51²-52²+……+97²-98²=(49+50)(49-50)+(51+52)(51-52)+……+(97+98)(97-98)=99*(-1)+103*(-1)+……+195*(-1)=-99-103-……-195=-(99+195)*(97-49)/2=-294*24=-7056

49²-50²=(49+50)(49-50)=-(49+50)
51²-52²=(51+52)(51-52)=-(51+52)
所以
49²-50²+51²-52²+……+97²-98²=-(49+50……+97+98)
=-50(49+98)/2=-3675
不懂可以追问,谢谢!

49²-50²+51²-52²+……+97²-98²
= (49²-50²) + (51²-52²) +……+ (97²-98²)
= (49-50)(49+50) + (51-52)(51+52) +……+ (97-98)(97+98)
= (-1)(49+50+51+52+ ...... + 97+98)
= (-1)(49+98)*50/2
= -3675

=(49+50)(49-50)+(51+52)(51-52)+...+(97+98)(97-98)
=-49-50-51-52-...-97-98
=-1(49+98)*(50/2)
=-3675