A bài thầy Phú

Ta có: N=1.99+2.98+3.97+..+98.2

=(99-98).99+(99-97).98+(99-96).97+..+(99-1).2

=99.99-98.99+98.99-97.98+99.97-96.97+..+99.2-1.2

=99.99+99.98+99.97+..+99.2-(99.98+98.97+97.96+..+1.2)

=99.(99+98+97+..+2)-323400 (cm a)

Xét dãy 99+98+97+..+2

SSH:(99-2):1+1=98

Tổng: (99+2).98:2=4949

=>N=99.4949-323400

=489951-323400=166551

Ta có:Xét tử số 

TS có 99 tổng,1 có mặt trong 99 tổng,2 có mặt trong 98 tổng,3 có mặt trong 97 tổng,...,99 có mặt trong 1 tổng

Vì thế ta được:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+1.99}\)

\(=\frac{1.99+2.98+3.97+...+99.1}{1.99+2.98+3.97+...+99.1}=1\)

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+...+98,2+99,1 thì sẽ bằng => 166 650

\(C=1.99+2.98+3.97+........+97.3+98.2+99.1\)

\(\Rightarrow C=1.99+2.\left(99-1\right)+3\left(99-2\right)+..........+98.\left(99-97\right)+99.\left(99-98\right)\)

\(\Rightarrow C=1.99+2.99-1.2+3.99-2.3+........+98.99-97.98+99.99-98.99\)

\(\Rightarrow C=\left(1.99+2.99+.......+99.99\right)-\left(1.2+2.3+.........+98.99\right)\)

\(\Rightarrow C=490050-\left(1.2+2.3+....+98.99\right)\)

Đặt \(A=1.2+2.3+3.4+........+98.99\)

\(\Rightarrow3A=1.2.3+2.3.3+..........+98.99.3\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+.....+98.99\left(100-97\right)\)

\(\Rightarrow3A=1.2.3-1.2.3+2.3.4-2.3.4+......+97.97.99-97.98.99+98.99.100\)

\(\Rightarrow3A=98.99.100\)

\(\Rightarrow A=\dfrac{98.99.100}{3}=323400\)

\(\Rightarrow C=490050-323400=166650\)

Vậy \(C=166650\)

Xem lại đề đi bạn. Đề có vấn đề?!

Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)

\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)

\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)

\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)

\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)

\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)

\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)

\(=49\cdot50\cdot67+2500=166650\)

Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)

\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)

\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)

\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)

\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)

\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)

\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)

Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)

\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)

\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)

\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)

\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)

\(=50\cdot101\cdot70=3500\cdot101=353500\)

Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)

\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)

\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)

\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)

\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)

\(=49\cdot33\cdot203=328251\)

\(a,A=1\cdot2+2\cdot3+...+98\cdot99\\ 3A=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot4\cdot3+...+98\cdot99\cdot3\\ 3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\left(5-2\right)+...+98\cdot99\left(100-97\right)\\ 3A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+3\cdot4\cdot5-...-97\cdot98\cdot99+98\cdot99\cdot100\\ 3A=98\cdot99\cdot100=970200\\ A=323400\)

\(b,B=1^2+2^2+3^3+...+98^2\\ B=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+98\left(99-1\right)\\ B=\left(1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\right)-\left(1+2+...+98\right)\\ B=323400-\left[\left(98+1\right)\left(98-1+1\right):2\right]\\ B=323400-4851=318549\\ c,C=1\cdot99+2\left(99-1\right)+3\left(99-2\right)+...+98\left(99-97\right)+99\left(99-98\right)\\ C=1\cdot99+2\cdot99-1\cdot2+3\cdot99-2\cdot3+...+98\cdot99-97\cdot98+99\cdot99-98\cdot99\\ C=99\left(1+2+...+99\right)-\left(1\cdot2+2\cdot3+...+98\cdot99\right)\\ C=99\left[\left(99+1\right)\left(99-1+1\right):2\right]-323400\\ C=490050-323400=166650\)

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:vv hỏi hoài z?

bị lỗi

a bn tự lm nha mk lm đỡ bn phần b

b=1.99+1.98+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+99.99-99.98

=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+...+98.99)

=99.(1+2+3+...+98+99)-(1.2+2.3+...+98.99)

=99.4950-(1.2+2.3+...+98.99)

=490050-(1.2+2.3+...+98.99)

b=1.2+2.3+...+98.99

3b=1.2.3+2.3.3+...+98.99.3

3b=1.2.3+2.3.(4-1)+...+98.99.(100-97)

3b=1.2.3+2.3.4+2.3.1+...+98.99.100+98.00.97

3b=490050-(98.99.100):3

b=490050-323400

b=166650

tk mk nha

=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