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\(C=1.99+2.98+3.97+...+99.1\)
\(=1.99+2\left(99-1\right)+3\left(99-2\right)+...+99\left(99-98\right)\)
\(=99\left(1+2+3+...+99\right)-\left(2+2.3+3.4+...+98.99\right)\)
\(=\frac{99\left(1+99\right).99}{2}-\frac{98.99.100}{3}\)
\(=99.50.99-98.33.100\)
\(=499050-323400\)
\(=166650\)
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
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)
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