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1.4+2.5+3.6+...+99.102=1(2+2)+2(3+2)+3(4+2)+...99(100+2)
=1.2+1.2+2.3+2.2+...+99.100+99.2
=(1.2+2.3+...+99.100)+2(1+2+...+99)
A=1.2+2.3+3.4+...+99.100(cho A la ten bieu thuc nay)
3A=1.2(3-0)+2.3(4-1)+...+99.100(101-98)
=(1.2.3+2.3.4+...+99.100.101)-(1.2.3+2.3.4+3.4.5+...+98.99.100
=99.100.101=>A=\(\frac{99.100.101}{3}\)=33330
2.(1+2+...99)
=2(100.99:2)=2.4950=9900
33330+9900=343200
vay...
D =1.4+2.5+3.6+.......+99.102
D = 1. (2+2) +2.(2+3) +3.(2+4)+...+99.(100+2)
D = 1.2+1.2+2.2+2.3+2.3+3.4+...+2.99+99.100
D = (1.2+2.3+3.4+...+99.100) +2.(1+2+3+4+...+99)
*Gọi A= 1.2+2.3+3.4+...+99.100
3A = 3.(1.2+2.3+3.4+...+99.100)
3A = 1.2.3+2.3.3+...+99.100.3
3A = 1.2.3 +2 .3.(4-1)+...+99.100.(101-98)
3A = 1.2.3+2.3.4-1.2.3+...+ 99.100.101-98.99.100
3A = 99.100.101
3A = 3.33.100.101
A = 33.100.101
A = 333300
* Gọi B = 2. (1+2+3+4+...+99)
\__có 99 số hạng ___/
B= 2.[(1+99).99:2]
B = 2 .4950
B = 9900
A+B = 333300+9900 =343200
Vậy D =343200
Ta có: A=1.4+2.5+3.6+.....+98.101+99.102
A=1.(2+2)+2.(3+2)+3(4+2)+........+98.(100+1)+99(100+2)
A=1.2+1.2+2.3+2.2+3.4+3.2+.........+98.100+98.1+99.100+99.2
A=(1.2+2.3+3.4+.......+99.100)+2.(1+2+3+4+......+99)
A=...................
1.4+2.5+3.6+...+99.102
=1(2+2)+2(3+2)+3(4+2)+...99(100+2)
=1.2+1.2+2.3+2.2+...+99.100+99.2
=(1.2+2.3+...+99.100)+2(1+2+...+99)
A=1.2+2.3+3.4+...+99.100(cho A la ten bieu thuc nay)
3A=1.2(3-0)+2.3(4-1)+...+99.100(101-98)
=(1.2.3+2.3.4+...+99.100.101)-(1.2.3+2.3.4+3.4.5+...+98.99.100
=99.100.101=>A=99.100.101399.100.1013=33330
2.(1+2+...99)
=2(100.99:2)=2.4950=9900
33330+9900=343200
Vậy B = 343200
\(B=1.4+2.5+3.6+...+99.102=1\left(2+2\right)+2\left(3+2\right)+3\left(4+2\right)+...+99\left(100+2\right)\)
\(=1.2+1.2+2.3+2.2+3.4+3.2+...+99.100+99.2\)
\(=\left(1.2+2.3+3.4+...+99.100\right)+2.\left(1+2+3+...+99\right)\)
Tính \(E=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3E=1.2.\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+99.100\left(101-98\right)\)\(=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(1.2.3+2.3.4+3.4.5+...+98.99.100\right)\)
\(=99.100.101\Rightarrow E=\frac{99.100.101}{3}=333300\)
Tính \(F=2.\left(1+2+3+...+99\right)\)
\(=2.\left(\frac{\left(99+1\right)99}{2}\right)=100.99=9900\)
Vậy, \(B=E+F=333300+9900=343200\)
https://olm.vn/hoi-dap/question/271984.html
\(A=1.4+2.5+3.6+....+99.102\)
\(A=1.\left(1+3\right)+2.\left(2+3\right)+3.\left(3+3\right)+...+99\left(99+3\right)\)
\(A=1^2+1.3+2^2+2.3+3^2+3.3+....+99^2+99.3\)
\(A=\left(1^2+2^2+......+99^2\right)+\left(1.3+2.3+.....+99.3\right)\)
\(A=\left(1^2+2^2+...+99^2\right)+3\left(1+2+....+99\right)\)
Làm tiếp nhé :(
\(A = 1.4 + 2.5 + 3.6 + ...+ 99.102\)
\(A=1.2+1.2+2.3+2.2+3.4+3.2+...+99.100+99.2\)
\(A=(1.2+2.3+3.4+...+99.100)+2.(1+2+3+...+99)\)
\(A=333300+9900\)
\(A=343200\)
\(B = 2.4 + 4.6 + 6.8 + ....+ 98.100 + 100.102\)
\(B=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2) \)
\(B=4(1.2+2.3+...+50.51) \)
\(M= 1.2+2.3+...+50.51 \)
\(3M=1.2.3+2.3.(4-1)+...+50.51.(52-49) \)
\(=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51 \)
\(= 50.51.52\)
\(=132600 \)
\(\Rightarrow\)\(M=44200 \)
\(\Rightarrow\) \(B=4M=176800\)
nguyen trung nghia
A = 1.4 + 2.5 + 3.6 + ... + 99.102
A = 1.(2 + 2) + 2.(3 + 2) + 3.(4 + 2) + ... + 99.(100 + 2)
A = (1.2 + 2.3 + 3.4 + ... + 99.100) + (1.2 + 2.2 + 3.2 + ... + 99.2)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100
3B = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 99.100.(101-98)
3B = 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3B = 99.100.101
B = 33.100.101 = 333300
A = 333300 + 2.(1 + 2 + 3 + ... + 99)
A = 333300 + 2.(1 + 99).99:2
A = 333300 + 100.99
A = 333300 + 9900
A = 343200
A = 1.( 2 + 2 ) + 2.( 3 + 2 ) + 3.( 4 + 2 ) + ... + 99.( 100 + 2 )
A = 1.2 + 1.2 + 2.2 + 2.3 + 3.4 + 3.2 +... + 99.100 + 99.2
A = ( 1.2 + 2.3 + 3.4 + ... + 99.100 ) + ( 1 + 2 + 3 + ... + 99 )
A = 333300 + 9900
A = 343200