Question

Tính: B= 1x2+2x3+3x4+...+2009x2010

 

Answer 1

 

B= 1.2+2.3+3.4+...+2009.2010

=>3B=1.2.3+2.3.3+3.4.3+...+2009.2010.3

=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+2009.2010.(2011-2008)

=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+....+2009.2010.2011-2008.2009.2010

=2009.2010.2011

=>B=\(\frac{2009.2010.2011}{3}=2706866330\)

 

Answer 2

ta có: 1x2+2x3+3x4+....+n(n+1) 
=1x(1+1)+2x(2+1)+3x(3+1)+....­n(n+1) 
=(1^2+2^2+3^2+¡­+n^2)+(1+2+3+....+n) 
=n(n+1)(2n+1)/6+n(n+1)/2 
=[n(n+1)[(2n+1)+3]/6

thay n=2009=> B=\(\frac{2009.\left(2009+1\right).\left(2009.2+1\right)+3}{6}\)=2704847286