Question

1²－2²＋3²－4²＋……＋99²－100²＝

简便计算

Answer 1

每两个为一组，利用平方差公式的逆式计算
1²-2²+3²-4²+5²-6²+…+99²-100²
=(1-2)(1+2)+(3-4)(3+4)+…+(99-100)(99+100)
=-(1+2+3+4+…+99+100)
=-5050

结果为负5050

Answer 2

an
=n^2
=n(n+1) -n
=(1/3)[n(n+1)(n+2) - (n-1)n(n+1)] - (1/2)[ n(n+1) -(n-1)n]
Sn
= a1+a2+...+an
=(1/3)n(n+1)(n+2) - (1/2)n(n+1)
=(1/6)n(n+1) [ 2(n+2) - 3]
=(1/6)n(n+1)(2n+1)
1^2-2^2+3^2-4^2+....+99^2 -100^2

=(1^2+2^2+....+100^2) -2( 2^2+4^2+...+100^2)
=(1^2+2^2+....+100^2) -8( 1^2+2^2+...+50^2)
=S100 -8S50
=(1/6)(100)(101)(201) -(8/6)(50)(51)(101)
=-5050

Answer 3

供参考。