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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
Lời giải:
$(1.2+2.3+3.4+...+2012.2013)-(2^2+3^2+...+2013^2)$
$=[(2-1).2+(3-1).3+(4-1).4+...+(2013-1).2013]-(2^2+3^2+...+2013^2)$
$=(2^2+3^2+4^2+...+2013^2)-(2+3+4+...+2013)-(2^2+3^2+...+2013^2)$
$=-(2+3+4+...+2013)$
$=1-(1+2+3+...+2013)$
$=1-2013.2014:2=1-2027091=-2027090$
(1.2+2.3+3.4+.....+2012.2013)-(22+32+42+......+20132)
= 1.2 + 2.3 + 3.4 +...+ 2012.2013 - 22 -32 - 42 -....-20132
=1.2 + 2.3 + 3.4 + ...+2012.2013 - 2.2 -3.3 - 4.4 -...- 2013.2013
=(1.2 - 2.2) + (2.3 - 3.3) + (3.4 - 4.4) + ...+(2012.2013 - 2013.2013)
=2.(1-2) + 3.(2-3) + 4.(3-4) +...+2013.(2012-2013)
=2.(-1) + 3.(-1) + 4.(-1) + ...+2013.(2012-2013)
= -2 - 3 - 4 -...- 2013
= -(2+3+4+...+2013)
= -[(2013+2).2012:2]
=-2027090
\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
1.Thực hiện phép tính một cách hợp lý:
a,(1.2+2.3+3.4+...+2012.2013)-(\(^{2^2+3^2+4^2+...+2013^2}\))
(1.2 + 2.3 + 3.4 + ... + 2012.2013) - (22 + 32 + 42 + 52 + ... + 20132)
= [(2 - 1).2 + (3 - 1).3 + (4 - 1).4 + ... + (2013 - 1).2013] - (22 + 32 + 42 + 52 + ... + 20132)
= (22 - 2 + 32 - 3 + 42 - 4 + ... + 20132 - 2013) - (22 + 32 + 42 + 52 + ... + 20132)
= 22 - 2 + 32 - 3 + 42 - 4 + ... + 20132 - 2013 - 22 - 32 - 42 - 52 - ... - 20132
= (22 - 22) + (32 - 32) + (42 - 42) + ... + (20132 - 20132) - (2 + 3 + 4 + ... + 2013)
= 0 - (2 + 3 + 4 + ... + 2013)
= 0 - (1 + 2 + 3 + 4 + ... + 2013) + 1
= 0 - \(\dfrac{2013.\left(2013+1\right)}{2}\) + 1
= 0 - 2027091 + 1
= (-2027091) + 1
= -2027090