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BK
1
DP
26 tháng 6 2017
Bài 3 :
\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{99\times100}\)
Ta có : \(\frac{1}{1\times2}=\frac{2-1}{1\times2}=\frac{2}{1\times2}-\frac{1}{1\times2}=1-\frac{1}{2}\)
\(\frac{1}{2\times3}=\frac{3-2}{2\times3}=\frac{3}{2\times3}-\frac{2}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{99\times100}=\frac{100-99}{99\times100}=\frac{100}{99\times100}-\frac{99}{99\times100}=\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10\times11}+\frac{1}{11\times12}+...+\frac{1}{38\times39}\)
Ta có : \(\frac{1}{10\times11}=\frac{11-10}{10\times11}=\frac{11}{10\times11}-\frac{10}{10\times11}=\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{11\times12}=\frac{12-11}{11\times12}=\frac{12}{11\times12}-\frac{11}{11\times12}=\frac{1}{11}-\frac{1}{12}\)
\(\frac{1}{38\times39}=\frac{39-38}{38\times39}=\frac{39}{38\times39}-\frac{38}{38\times39}=\frac{1}{38}-\frac{1}{39}\)
\(\frac{1}{39\times40}=\frac{40-39}{39\times40}=\frac{40}{39\times40}-\frac{39}{39\times40}=\frac{1}{39}-\frac{1}{40}\)
\(\Rightarrow B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
ML
26 tháng 6 2017
3.
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{38.39}+\frac{1}{39.40}\)
\(B=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{38}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}\)
\(B=\frac{1}{10}-\frac{1}{40}\)
\(B=\frac{3}{40}\)
CM
12 tháng 9 2017
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
16 tháng 11 2021
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CM
18 tháng 6 2018
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 6 2023
1 \(\times\) 2 \(\times\) 3 = 1 \(\times\) 2 \(\times\) 3
2 \(\times\) 3 \(\times\) 3 = 2 \(\times\) 3 \(\times\) ( 4 -1) = 2 \(\times\) 3 \(\times\) 4 - 1 \(\times\) 2 \(\times\) 3
3 \(\times\) 4 \(\times\) 3 = 3 \(\times\) 4 \(\times\) ( 5 -2) = 3 \(\times\) 4 \(\times\) 5 - 2 \(\times\) 3 \(\times\) 4
4 \(\times\) 5 \(\times\) 3 = 4 \(\times\) 5 \(\times\) ( 6- 3) = 4 \(\times\) 5 \(\times\) 6 - 3 \(\times\) 4 \(\times\) 5
..................................................................................
99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)(101-98) =99\(\times\)100\(\times\)101 - 98\(\times\)99\(\times\)100
Cộng vế với vế ta được:
1\(\times\)2\(\times\)3 + 2\(\times\)3\(\times\)3 + 3\(\times\)4\(\times\)3+ ...+99\(\times\)100\(\times\)3 = 99\(\times\)100\(\times\)101
(1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4 +...+99\(\times\)100)\(\times\)3 = 99\(\times\)100\(\times\)101
1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = (99 \(\times\)100 \(\times\)101):3
1\(\times\)2 + 2\(\times\)3 + 3\(\times\)4+...+99\(\times\)100 = 333 300
HN
12
CM
1 tháng 2 2017
A= 1x2+2x3+3x4+...+98x99 A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97) = 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97) = 98x99x100
CM
22 tháng 6 2019
A= 1x2+2x3+3x4+...+98x99
A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97)
= 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97)
= 98x99x100.
16 tháng 10 2021
\(=\dfrac{1}{3}\left(1\times2\times3+2\times3\times3+...+19\times20\times3\right)\\ =\dfrac{1}{3}\left[1\times2\times\left(3-0\right)+2\times3\times\left(4-1\right)+...+19\times20\times\left(21-18\right)\right]\\ =\dfrac{1}{3}\left(1\times2\times3-1\times2\times3+2\times3\times4-...-18\times19\times20+19\times20\times21\right)\\ =\dfrac{1}{3}\times19\times20\times21=2660\)
Đặt A = 1.2 + 2.3 + 3.4 + .... + 39.40
3A = 3( 1.2 + 2.3 + 3.4 + .... + 39.40 )
= 1.2.3 + 2.3.3 + 3.4.3 + .... + 39.40.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 39.40(41 - 38)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + 39.40.41 - 38.39.40
= 39.40.41
\(\Rightarrow A=\frac{39.40.41}{3}=21320\)