Question

1*2+2*3+3*4+4*5+.+2002*2003是几

Answer 1

第n项是n×(n+1)=n²+n

所以原式=(1²+1)+(2²+2)+……+(2002²+2003)

=(1²+2²+3²+……+2002²)+(1+2+……+2003)

=2002×(2002+1)×(2×2002+1)/6 + 2003×(2003+1)/2

=2678686011