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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
Ta có :
C = 1.99 + 2.(99 - 1) + 3.(99 - 2) + ... + 98.(99 - 97) + 99.(99 - 98)
C = 99.(1 + 2 + 3 + ... + 98 + 99) - (2 + 2.3 + 3.4 + ...+97.98 + 98.99)
C = 99.(1 + 99).99/2 - 98.99.100/3
C = 99.50.99 - 98.33.100
C = 490050 - 323400
C = 166650
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
S = (50 -49)(50 +49) + (50 - 48)(50 + 48) + .... + (50 + 48)(50 -48)
S = 502 - 492 + 502 - 482 + ..... + 502 - 482 + 502 - 49
S = 99 . 502 - 2.(02 + 12 +.... + 492)
\(A = 1.99 + 2.98 + 3.97 + ...+ 97.3 + 98.2 + 99.1\)
\(A=1.99+2.\left(99-1\right)+3.\left(99-2\right)+...+98.\left(99-97\right)+99.\left(99-98\right)\)
\(A=1.99+2.99-1.2+3.99-2.3+98.99-97.98+99.99-98.99\)
\(=\left(1.99+2.99+3.99+...+98.99+99.99\right)-\left(1.2+2.3+3.4+...+97.98+98.99\right)\)
\(=99.\left(1+2+3+...+98+99\right)-\left(1.2+2.3+3.4+...+97.98+98.99\right)\)
\(=99.4950-\left(1.2+2.3+3.4+97.98+98.99\right)\)
Mà \(1.2+2.3+3.4+...97.98+98.99\)
\(=\frac{1}{3}.\left[1.2+2.3.\left(4-1\right)+3.4.\left(5-2\right)+98.99.\left(100-97\right)\right]\)
\(=\frac{1}{3}.98.99.100=323400\)
\(\Rightarrow A=99.4950-323400=166650\)
=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99(99-98)
=99.(1+2+3+4+...+98+99)-(2+2.3+3.4+...+97.98+98.99)
=99.(1+99).99/2-98.99.100/3
=99.50.99-98.33.100
=490050-323400
=166650