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c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4
==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)
==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11
==> 4C= 8.9.10.11=7920
==> C= 7920 :4=1980
a, Ta có: 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+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)
3A= 99.100.101 - 0.1.2
3A= 999900 - 0
3A= 999900
==> A= 999900 : 3
==> A= 333300
. mỗi hạng tử của tổng A có hai thừa số thì ta nhân A với 3 lần khoảng cách giữa hai thừa số đó. Häc tËp c¸ch ®ã , trong bài này ta nhân hai vế của A với 4 lần khoảng cách đó vì ở đây mỗi hạng tử có 3 thừa số .Ta giải được bài toán nh sau :
A = 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + 5.6.7 + 6.7.8 + 7.8.9 + 8.9.10
4A = (1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + 5.6.7 + 6.7.8 + 7.8.9 + 8.9.10).4
4A = [1.2.3.(4 – 0) + 2.3.4.(5 – 1) + ... + 8.9.10.(11 – 7)]
4A = (1.2.3.4 – 1.2.3.4 + 2.3.4.5 – 2.3.4.5 + ... + 7.8.9.10 – 7.8.9.10 + 8.9.10.11) 4A = 8.9.10.11 = 1980.
Tõ ®ã ta có kết quả tổng quát
A = 1.2.3 + 2.3.4 + 3.4.5 + ... + (n – 1).n.(n + 1).= (n -1).n.(n + 1)(n + 2)/4
\(A=\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+\dfrac{1}{7.8.9}+\dfrac{1}{8.9.10}\)
\(\Rightarrow2A=\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}+\dfrac{2}{7.8.9}+\dfrac{2}{8.9.10}\)
\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Rightarrow2A=\dfrac{1}{2.3}-\dfrac{1}{9.10}\)
\(\Rightarrow2A=\dfrac{22}{45}\)
\(\Rightarrow A=\dfrac{11}{45}\)
1/2x3x4 + 1/3x4x5 + 1/4x5x6 + 1/5x6x7 + ..... + 1/8x9x10
= { 2/2x3x4 + 2/3x4x5 + 2/4x5x6 + .... + 2/8x9x10 } : 2
= { 4-2/2x3x4 + 5-3/3x4x5 + 6-4/4x5x6 + .... + 10-8/8x9x10 } : 2
= { 4/2x3x4 - 2/2x3x4 + 5/3x4x5 - 3/3x4x5 + ... + 10/8x9x10 - 8/8x9x10 } : 2
= { 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ... + 1/8x9 - 1/9x10 } : 2
= { 1/2x3 - 1/9x10 } :2
= { 1/6 - 1/90 } : 2
= 14/90 : 2
= 7/90
c) \(C=1.2+2.3+3.4+...+98.99\)
\(\Rightarrow3C=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+98.99\left(100-97\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(=98.99.100\)
\(\Rightarrow C=\frac{98.99.100}{3}=323400\)
d) \(D=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
Ta có A = \(\frac{1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8}{2.4.6-4.6.8+6.8.10-8.10.12+10.12.14-12.14.16}\)
A = \(\frac{1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8}{\left(1.2.3\right).2-\left(2.3.4\right).2+\left(3.4.5\right).2-\left(4.5.6\right).2+\left(5.6.7\right).2-\left(6.7.8\right).2}\)
A = \(\frac{1.\left(1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8\right)}{2.\left(1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8\right)}\)
A = \(\frac{1}{2}\)
\(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(3A=99\cdot100\cdot101\Rightarrow A=\dfrac{99\cdot100\cdot101}{3}=333300\)
\(B=1^2+2^2+3^2+...+99^2+100^2\)
\(=\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=\dfrac{2030100}{6}=338350\)
\(C=1\cdot2\cdot3+2\cdot3\cdot4+...+8\cdot9\cdot10\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+8\cdot9\cdot10\cdot\left(11-7\right)\)
\(4C=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+8\cdot9\cdot10\cdot11-7\cdot8\cdot9\cdot10\)
\(4C=8\cdot9\cdot10\cdot11\Rightarrow C=\dfrac{8\cdot9\cdot10\cdot11}{4}=1980\)
@Hồng Phúc Nguyễn