bo tay

Ô tô đi với vận tốc 50km/giờ vì :

         100 : 2 = 50

                   đs : 50

Ta có: C = 1.2 + 2.3 + 3.4 + ..... + 98.99

=> 3C = 3.(1.2 + 2.3 + 3.4 + ..... + 98.99)

=> 3C = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 98.99.(100 - 97)

=> 3C = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100

=> 3C = 98.99.100

=> C = \(C=\frac{98.99.100}{3}=485100\)

Ô tô đi với vận tốc 50km/giờ vì :

         100 : 2 = 50

                   đs : 50

Ta có: C = 1.2 + 2.3 + 3.4 + ..... + 98.99

=> 3C = 3.(1.2 + 2.3 + 3.4 + ..... + 98.99)

=> 3C = 1.2.(3 - 0) + 2.3.(4 - 1) + .... + 98.99.(100 - 97)

=> 3C = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100

=> 3C = 98.99.100

=> C = $C=\frac{98.99.100}{3}=485100$

Trả lời hộ mình câu h= 1.1+2.2+ 3.3+4.4+...+98.98

A=3+3^2+3^3+...+3^2004
Ta có:A=(3+3^2+3^3+3^4)+...+(3^2001+3^2002+3^2003+3^2004)
=>A=120+...+(3^2000.3+3^2000.3^2+3^2000.3^3+3^2000.3^4)
=>A=120+...+3^2000(3+3^2+3^3+3^4) 
=>A=120+...+3^2000.120
=>A=(1+...+3^2000).120
Vì 120 chia hết cho 120 nên A chia hết cho 120=>A chia hết cho 10
A=3+3^2+3^3+...+3^2004
=>A=(3+3^2+3^3)+...+(3^2002+3^2003+3^2004)
=>A=39+...+(3^2000.3+3^2000.3^2+3^2000.3^3)
=>A=39+...+3^2000(3+3^2+3^3)
=>A=39+...+3^2000.39
=>A=(1+...+3^2000).39
Vì 39 chia hết cho 13 nên A chia hết cho 13
Ta có:A chia hết cho 10;A chia hết cho 13 và (10;13)=1 nên A chia hết cho 10.13
=>A chia hết cho 130
 Vậy...

Đặt A=1.2+2.3+3.4+............+1999.2000

3A=1.2.3+2.3.(4-1)+.................+1999.2000.(2001-1998)

3A=1.2.3+2.3.4-1.2.3+............+1999.2000.2001-1998.1999.2000

3A=1999.2000.2001

A=1999.2000.2001:3

A=2666666000

b,Đặt B=1.2+2.2+3.3+............+1999.1999

B=1.(2-1)+2.(3-1)+3.(4-1)+..........+1999.(2000-1)

B=1.2-1+2.3-2+3.4-3+...........+1999.2000-1999

B=(1.2+2.3+3.4+.............+1999.2000)-(1+2+3+...........+1999)

B=2666666000-1999000

B=2664667000

c,Đặt C=1.2.3+2.3.4+..........+48.49.50

4C=1.2.3.4+2.3.4.(5-1)+.........+48.49.50.(51-47)

4C=1.2.3.4+2.3.4.5-1.2.3.4+..............+48.49.50.51-47.48.49.50

4C=48.49.50.51

C=48.49.50.51:4

C=1499400

b)Ta chứng minh công thức \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\) (*)

Với n=1 (*) đúng

Giả sử (*) đúng với n=k, khi đó ta có

\(1^2+2^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)

Ta chứng minh (1) đúng với n=k+1, từ (1) suy ra:

\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)

\(=\left(k+1\right)\left(\frac{k\left(2k+1\right)}{6}+k+1\right)=\left(k+1\right)\frac{2k^2+7k+6}{6}\)

\(=\frac{\left(k+1\right)\left(2k^2+4k+3k+6\right)}{6}=\frac{\left(k+1\right)\left[2k\left(k+2\right)+3\left(k+2\right)\right]}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

Theo nguyên lí quy nạp ta có ĐPCM

Áp dụng vào bài toán ta có:

\(B=\frac{98\left(98+1\right)\left(2\cdot98+1\right)}{6}=318549\)

a)\(A=1\cdot2+2\cdot3+...+98\cdot99\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+98\cdot99\left(100-97\right)\)

\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+98\cdot99\cdot100-97\cdot98\cdot99\)

\(3A=98\cdot99\cdot100=\frac{98\cdot99\cdot100}{3}=323400\)

câu c thì vừa này t vào xem phần hỏi đáp trang toán có bài đăng r` đấy quay lại xem

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

không biết

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

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

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

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

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

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?!

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

\(=1.99+2(99-1)+3(99-2)+...+97.(99-96)+98(99-97)+99(99-98)\)

\(=(1.99+2.99+3.99+...+97.99+98.99+99.99)-(1.2+2.3+...+96.97+97.98+98.99)\)

\(=99.(1+2+3+...+99)- (1.2+2.3+...+96.97+97.98+98.99)\)

\(=99.\frac{99.(99+1)}{2} - (1.2+2.3+...+96.97+97.98+98.99)\)

\(=490050 - (1.2+2.3+...+96.97+97.98+98.99)\)

Từ đây tính vế trong ngoặc ra rồi thực hiện trừ là oke

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

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

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