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S=1x2+2x3+3x4+4x5+...+98x99
3S= 1.2.3+ 2.3.3 + 3.4.3 + 4.5.3+...+98.99.3
3S= 1.2.3+ 2.3(4-1) + 3.4(5-2) + 4.5(6-3)+....+ 98.99.(100-97)
3S= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5 - 2.3.4 +...+98.99.100 -97.98.99
3S= 98.99.100
S=970200:3
S= 323400
Bài làm:
\(S=1.2+2.3+3.4+...+98.99\)
\(S=\frac{1}{3}\left(1.2.3+2.3.3+3.4.3+...+98.99.3\right)\)
\(S=\frac{1}{3}\left[1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\right]\)
\(S=\frac{1}{3}\left(1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-97.98.99+98.99.100\right)\)
\(S=\frac{98.99.100}{3}=323400\)
Vậy S = 323400
Học tốt!!!!
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
easy mà
=2.(1/1.2+1/2.3+...+1/1999.2000)
=2.(1/1-1/2+1/2-1/3+....1/1999-1/2000)
=2.(1-1/2000)
=2.1999/2000
=3998/2000=... tự rút gọn :D
S = 1x2 + 2x3 + 3x4 + ……………… + 11x12 + 12x13
3S=1x2x3 + 2x3x3 + 3x4x3+ ………. + 11x12x3 + 12x13x3
Ta lấy K = 1x2x3 +2x3x4 + 3x4x5 + …… + 11x12x13 + 12x13x14
- 3S = 1x2x3 + 2x3x3 + 3x4x3+ ……… + 11x12x3 + 12x13x3
------------------------------------------------------------------------------------
K – 3S = 0 + 2x3x1 + 3x4x2 + …… .. + 11x12x10 + 12x13x11
K – 3S = K – 12x13x14
Từ đó suy ra: 3S = 12x13x14
S = 4x13x14 = 728
Cách 2:
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + …. + 11x12x(13-10) + 12x13x(14-11)
S x 3 = 1x2x3 + 2x3x4 – 2x3x1 + 3x4x5 – 3x4x2 + …..+ 11x12x13 – 11x12x10 +12x13x14 – 12x13x11
S x 3 = 12 x 13 x14
S = 4 x 13 x 14
S = 728
ai k minh minh k lai cho
S= 2x(1/1x2+1/2x3+1/3x4+...........+1/2020x2021)
S=2x(1-1/2+1/2-1/3+1/3-...+1/2020-1/2021)
S=2x(1-1/2021)
S=2x2020/2021
S=4040/2021
2019/2010<3/2<4040/2021
=>2019/2010<S
S = 2 x (\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\)\(\frac{2}{2020\times2021}\))
= 2 x (\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\)\(\frac{1}{2020\times2021}\))
= 2 x ( \(1-\frac{1}{2021}\))
= \(2\times\frac{2020}{2021}\)
= \(\frac{4040}{2021}\)
= \(\frac{4042-2}{2021}\)
\(=2-\frac{2}{2021}\)
Ta có :
\(\frac{2019}{2010}=\frac{2020-1}{2010}=2-\frac{1}{2010}=2-\frac{2}{2020}\)
Ta thấy \(\frac{2}{2021}< \frac{2}{2020}\)
nên \(2-\frac{2}{2021}>2-\frac{2}{2020}\)
Vậy \(S\)\(>\frac{2019}{2010}\)