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\(A=1\cdot2+2\cdot3+...+151\cdot152\)
\(=1\left(1+1\right)+2\left(1+2\right)+...+151\left(1+151\right)\)
\(=\left(1+2+3+...+151\right)+\left(1^2+2^2+...+151^2\right)\)
\(=\dfrac{151\left(151+1\right)}{2}+\dfrac{151\left(151+1\right)\left(2\cdot151+1\right)}{6}\)
\(=151\cdot76+\dfrac{151\cdot152\cdot303}{6}\)
\(=151\cdot76+151\cdot7676=1170552\)
\(C=2\cdot4+4\cdot6+...+2024\cdot2026\)
\(=2\cdot2\left(1\cdot2+2\cdot3+...+1012\cdot1013\right)\)
\(=4\left[1\left(1+1\right)+2\left(1+2\right)+...+1012\left(1+1012\right)\right]\)
\(=4\left[\left(1+2+...+1012\right)+\left(1^2+2^2+...+1012^2\right)\right]\)
\(=4\left[1012\cdot\dfrac{1013}{2}+\dfrac{1012\left(1012+1\right)\left(2\cdot1012+1\right)}{6}\right]\)
\(=4\left[506\cdot1013+345990150\right]\)
\(=1386010912\)
\(M=1^2+2^2+...+2024^2\)
\(=\dfrac{2024\left(2024+1\right)\cdot\left(2\cdot2024+1\right)}{6}\)
\(=2024\cdot2025\cdot\dfrac{4049}{6}\)
=2765871900
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
vậy thì tổng của : -1+(-2)+(-3)+.........+(-49) = -(1+2+3+..........+49) = -1225
333...3x666...6=333...3x(3x222...2)=999...9x222...2=(1000...0-1)x222...2=1000...0x222...2-222...2=222...2000...0-222...22
A = 1.2 + 2.3 + 3.4 + ... + 2013.2014
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2013.2014.3
Mà :
1.2.3 = 1.2.3
2.3.3 = 2.3.4 - 2.3.1
3.4.3 = 3.4.5 - 3.4.2
2012.2013.3 = 2012.2013.2014 - 2012.2013.2011
2013.2014.3 = 2013.2014.2015 - 2013.2014.2012
Cộng tất cả, vế theo vế ---> 3S = 2013.2014.2015
=> A = 2013.2014.2015 / 3 = 2723058910
1.
A = (22.21.20 - 2.1.0) : 3
A = 9240 : 3
A = 3080
3.A = 3080 x 3
3.A = 9240
Đặt S=1.2+2.3+.........+2011.2012
3S=1.2.3+2.3.(4-1)+...........+2011.2012.(2013-2010)
3S=1.2.3+2.3.4-1.2.3+...........+2011.2012.2013-2010.2011.2012
3S=2011.2012.2013
S=2011.2012.2013:3
S=2714954572
Đặt A = 1.2 + 2.3 + 3.4 + ... + 2011.2012
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2011.2012.3
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2011.2012.(2013 - 2010)
=> 3A = 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2011.2012.2013 - 2010.2011.2012
=> 3A = 2011.2012.2013
=> A = \(\frac{2011.2012.2013}{3}=2714954572\).
Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 49.50
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 49.50.51
=> 3A = 49.50.51
= >A = 49.50.51/3 = 41650
\(B=1.2+2.3+3.4+...+49.50\)
\(\Rightarrow3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(\Rightarrow3B=49.50.51\)
\(\Rightarrow B=\frac{49.50.51}{3}\)
B=\(1.2+2.3+....+49.50\\ \Rightarrow3B=1.2.\left(3-0\right)+2.3.\left(4-1\right)+.....+49.50.\left(51-48\right)\)
\(\Rightarrow3B=1.2.3-0.1.2+2.3.4-1.2.3.+.....+49.50.51-48.49.50\\ \Rightarrow3B=49.50.51\\ \Rightarrow3B=124950\\ \Rightarrow B=41650\)
C=\(1^2+2^2+3^2+....+50^2\\ =1.1+2.2+3.3+.....+50.50\\ =1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+....+50\left(51-1\right)\\ \)
\(=\left(1.2+2.3+3.4+.....+50.51\right)-\left(1+2+3+....+50\right)\)
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