Question

用简便方法计算1-2²+3²-4²+5²-6²+······+99²-100²

Answer 1

(1-2²)+(3²-4²)+(5²-6²)+······+(99²-100²)
=∑((2n-1)²-(2n)²), n=1,2,3,4,.....,50
=∑(1-4n), n=1,2,3,4,.....,50
=((1-4*1)+(1-4*50))*50/2
=(-3-199)*50/2
=-202*50/2
=-5050

Answer 2

用简便方法计算1-2²+3²-4²+5²-6²+······+99²-100²
解：原式=（3²-2²）+（5²-4²）+........+（99²-98²）+1-100²
=（3-2）×（3+2）+（5-4）×（5+4）+........+（99-98）（99+98）+1-100²
=5+9+...+197+1-100²
=（197+5）×49÷2+1-100²
=101×49+1-10000
=（100+1）×49+1-10000
=4949+1-10000
=-5050

Answer 3

(1-2²)+(3²-4²)+(5²-6²)+······+(99²-100²)
=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+······+(99-100)(99+100)
=-3-7-11-······-199
=-50×202/2=-5050