1.99^2+2.98^2+3.97^2+....+98.2^2+99.1^2 Giúp mình vớiiiii
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HN
4
NP
0
KN
3
8 tháng 6 2017
=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
18 tháng 9 2018
=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
I
2
17 tháng 5 2015
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
MT
17 tháng 5 2015
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
BN
14 tháng 6 2016
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
QD
14 tháng 6 2016
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
6 tháng 8 2016
C=1.99+2.98+3.97+98.2+99.1
C=(1x1x99x99)+(2x98x98x2)+3x97
C=9801+9604+291
C=19405+291
C=19696
VD
0
TN
3
31 tháng 1 2016
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
Ta có: \(1\cdot99^2+2\cdot98^2+\cdots+99\cdot1^2\)
\(=99^2\left(100-99\right)+98^2\left(100-98\right)+\cdots+1^2\left(100-1\right)\)
\(=100\left(1^2+2^2+\cdots+99^2\right)-\left(1^3+2^3+\cdots+99^3\right)\)
\(=100\cdot\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}-\left(1+2+\cdots+99\right)^2\)
\(=100\cdot\frac{99\cdot100\cdot199}{6}-\left\lbrack99\cdot\frac{100}{2}\right\rbrack^2\)
\(=100\cdot33\cdot50\cdot199-\left(99\cdot50\right)^2\)
\(=5000\cdot33\cdot199-99^2\cdot50^2=8332500\)