Lời giải:

$(1.2+2.3+3.4+...+2012.2013)-(2^2+3^2+...+2013^2)$

$=[(2-1).2+(3-1).3+(4-1).4+...+(2013-1).2013]-(2^2+3^2+...+2013^2)$

$=(2^2+3^2+4^2+...+2013^2)-(2+3+4+...+2013)-(2^2+3^2+...+2013^2)$

$=-(2+3+4+...+2013)$

$=1-(1+2+3+...+2013)$

$=1-2013.2014:2=1-2027091=-2027090$

Ta có:

3M=1.2.3+2.3.3+3.4.3+...+2012.2013.3

3M=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2012.2013.(2014-2011)

3M=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2012.2013.2014-2011.2012.2013

3M=2012.2013.2014

3M=8157014184

M=8157014184:3

M=2719004728

\(a,A=1\cdot2+2\cdot3+...+98\cdot99\\ 3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+98\cdot99\cdot3\\ 3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\left(5-2\right)+...+98\cdot99\left(100-97\right)\\ 3A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+3\cdot4\cdot5-...-97\cdot98\cdot99+98\cdot99\cdot100\\ 3A=98\cdot99\cdot100=970200\\ A=323400\)

\(b,B=1^2+2^2+3^3+...+98^2\\ B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+98\left(99-1\right)\\ B=\left(1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\right)-\left(1+2+...+98\right)\\ B=323400-\left[\left(98+1\right)\left(98-1+1\right):2\right]\\ B=323400-4851=318549\\ c,C=1\cdot99+2\left(99-1\right)+3\left(99-2\right)+...+98\left(99-97\right)+99\left(99-98\right)\\ C=1\cdot99+2\cdot99-1\cdot2+3\cdot99-2\cdot3+...+98\cdot99-97\cdot98+99\cdot99-98\cdot99\\ C=99\left(1+2+...+99\right)-\left(1\cdot2+2\cdot3+...+98\cdot99\right)\\ C=99\left[\left(99+1\right)\left(99-1+1\right):2\right]-323400\\ C=490050-323400=166650\)

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