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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
\(N=1^3+2^3+...+100^3\)
\(=\left(1+2+3+...+100\right)^2\)
\(=\left[\dfrac{100\left(100+1\right)}{2}\right]^2\)
\(=\left[50\cdot101\right]^2=5050^2\)
\(Q=1^3+2^3+...+2024^3\)
\(=\left(1+2+3+...+2024\right)^2\)
\(=\left[\dfrac{2024\left(2024+1\right)}{2}\right]^2\)
\(=\left[1012\left(2024+1\right)\right]^2\)
\(=2049300^2\)
Ta có:
3S = 1.2.3 + 2.3.4 + 3.4.3 + ... + 99.100.3
3S = 1.2 ( 3 - 0 ) + 2.3. ( 4 - 1 ) + 3.4 . ( 5 - 2 )............... 99.100 . ( 101 - 98 )
3S = ( 1.2.3 + 2.3.4 + 3.4.5 + ... + 99.100.101 ) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 98.99.100 )
3S = 99.100.101 - 0.1.2
3S = 999900 - 0
3S = 999900
S = 999900 : 3
S = 333300
Gọi A là biểu thức ta có:
A = 1.2+2.3+3.4+......+99.100
Gấp A lên 3 lần ta có:
A . 3 = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
A . 3 = 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)
A . 3 = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100
A . 3 = 99.100.101
A = 99.100.101 : 3
A = 33.100.101
A = 333 300
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 = \(\frac{2011.2012.2013}{3}=2714954572\).
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3
=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 49.50.( 51 - 48 )
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 49.50.51 - 48.49.50
=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 48.49.50 - 48.49.50 ) + 49.50.51
=> 3A = 49.50.51
=> A = ( 49.50.51 ) : 3
=> A = 41650
A = 1.2 + 2.3 + 3.4 + ..... + 49.50
3A=1.2.3+2.3.3+3.4.3+...+49.50.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+48.49.(50-47)+49.50.(51-48)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50
3A=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...(47.48.49-47.48.49)-(48.49.50-48.49.50)+49.50.51
3A=0+0+...+0+0+49.50.51
3A=49.50.51
A=\(\frac{49.50.51}{3}\)
A=41650
Đáp số: A=41650
Ta có A=1.2+2.3+3.4+...+98.99
B=1^2+2^2+3^2+...+98^2=1.1+2.2+3.3+...+98.98
=> A-B=(1.2+2.3+3.4+...98.99)-(1.1+2.2+3.3+...+98.98)
=(1.2-1.1)+(2.3-2.2)+(3.4-3.3)+...+(98.99-98.98)
=1.(2-1)+2.(3-2)+3.(4-3)+...+98.(99-98)
=1.1+2.1+3.1+...98.1
=1+2+3+...+98=[98.(98+1)]/2=98.99/2=4851
đúng nha
Ta có:a=1.2+2.3+3.4+...+98.99
a=1(1+1)+2(2+1)+3(3+1)+...+98(98+1)
a=12+1+22+2+33+3+...+982+98
a=b+(1+2+3+4+...+98)
a=b+(1+98).98:2
a=b+4851
Vậy a-b =4851