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A+B = (1.99+2.98+3.97+...+99.1)+(1.101+2.102+3.103+...+99.199)
A+B = (1.99+1.101)+(2.98+2.102)+(3.97+3.103)+...+(99.1+99.199)
A+B = 1(99+101) + 2(98+102) + 3(97.103)+...+99(1+199)
A+B = 1.200 + 2.200 + 3.200 +...+ 99.200
A+B = 200.(1+2+3+...+200)
A+B = 200.4950
A+B = 990000
A + B = ( 1 . 99 + 2 . 98 + 3 . 97 + ... + 99 . 1 ) + ( 1 . 101 + 2 . 102 + 3 . 103 + ... + 99 . 199 )
A + B = 99 . ( 1 + 199 ) + 98 . ( 2 + 198 ) + 97 . ( 3 + 197 ) + ... + 2 . ( 102 + 98 ) + 1 . ( 99 + 101 )
A + B = 99 . 200 + 98 . 200 + 97 . 200 + ... + 2 . 200 + 1 . 200
A + B = ( 99 + 98 + 97 + ... + 2 + 1 ) . 200
A + B = 4950 . 200
A + B = 990000
A+B=(1.99+2.98+...+99.1)+(1.101+2.102+...+99.199)
=(1.99+1.101)+(2.98+2.102)+...+(99.1+99.199)
=1.(99+101)+2.(98+102)+...+99(1+199)
=200+2.200+...+99.200
=200.(1+2+3+4+...+99)
=200.4950
=.....
a) 3.A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5- 2) +...+n.(n+1).(n+2 - (n-1)) + ...+ 97.98.(99- 96) + 98.99.(100 - 97)
=> 3.A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+ 97.98.99 - 96.97.98 + 98.99.100 - 97.98.99
= 98.99.100
=> A = 98.99.100 : 3 = 323400
b) B gồm 99 số 1; 98 số 2;..; 2 số 98; 1 số 99
Có thể Viết lại B = 1 + (1+2) + (1+2+3) +...+ (1+2+3+...+98 + 99)
= \(\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{98.99}{2}=\frac{1.2+2.3+3.4+...98.99}{2}=\frac{A}{2}=\frac{323400}{2}=161700\)
A = 1.2 + 2.3 + 3.4 +......+ 98.99
=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ........ + 98.99.(100 - 97)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ........ + 98.99.100 - 97.98.99
=> 3A = (1.2.3 + 2.3.4 + 3.4.5 + ....... + 98.99.100) - (1.2.3 + 2.3.4 + ..... + 97.98.99)
=> 3A = 98.99.100
=> A = \(\frac{98.99.100}{3}=323400\)
A = 1.2 + 2.3 + 3.4 +..... + 99.100
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + … + 99.100.3
=> 3A = 1.2.(3-0) + 2.3.(4 - 1) + 3.4.(5 - 2) + … + 99.100. (101 - 98)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … +99.100.101-98.99.100
=> 3A = 98.99.100
=> A = 99.100.101/3
=> A = 33.100.101 = 333300
Ta có: \(A=1\cdot99+2\cdot98+\cdots+99\cdot1\)
\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)
\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot49\cdot\frac{50}{2}-\frac{49\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot49\cdot25-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot49\cdot25-49\cdot25\cdot33\right\rbrack+2500=2\cdot25\cdot49\left(100-33\right)+2500\)
\(=50\cdot49\cdot67+2500=166650\)
Sửa đề: \(B=1\cdot101+2\cdot102+\cdots+9\cdot109\)
\(=1\left(100+1\right)+2\left(100+2\right)+\cdots+9\left(100+9\right)\)
=100(1+2+...+9)+(\(1^2+2^2+\cdots+9^2\) )
\(=100\cdot9\cdot\frac{10}{2}+\frac{9\left(9+1\right)\left(2\cdot9+1\right)}{6}\)
\(=900\cdot5+\frac{9\cdot10\cdot19}{6}=4500+3\cdot5\cdot19=4500+15\cdot19\)
=4500+285
=4785
A+B
=166650+4785
=171435
1.99+2.98+3.97+...+98.2+99.1=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+98.99)
=99.(1+2+...+99)-(1.2+2.3+...+98.99)=99.4950-(1.2+2.3+...+98.99)=490050-(1.2+2.3+...+98.99)
đặt A=1.2+2.3+...+98.99
=>3A=1.2.3+2.3.3+...+98.99.3
=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3-1.2.3+2.3.4-2.3.4+...+97.98.99-97.98.99+98.99.100=98.99.100
=>A=98.99.100:3=323400
=>1.99+2.98+3.97+...+98.2+99.1=490050-323400=166650
1.99+2.98+3.97+4.96+...+98.2+99.1
=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+4.99+...+98.99+99.99)-(1.2+2.3+3.4+...+97.98+98.99)
=(1+2+3+4+...+98+99).99-(98.99.100)/3
={(99-1+1)/2}.100.99-(98.99.100)/3
=49,5.100.99-(98.99.100)/3
=4950.99-(98.99.100)/3
=4950.3.33-98.100.33
B=14850.33-9800.33
B=(14850-9800).33
B=5050.33
B=166650