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\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\right).1428+185.8\)
\(=\frac{2}{2}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185.8\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}\right).1428+1480\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right).1428+1480\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right).1428+1480\)
\(\frac{1}{2}.\frac{370}{741}.1428+1480\)
\(=\frac{185}{741}.1428+1480\)
\(=356,52+1480=1836,52\)
chỗ\(\frac{185}{741}.1428\)mk làm tròn số lun á nha
mk ko chắc tính đúng hay sai nha nhưng cách làm thì kiểu vậy
549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
a) A = 1.3 +2.4 + 3.5 +...+ 97.99 + 98.100
A = 1(2 + 1) + 2(3+1) + 3(4 + 1) +...+ 98(99+1)
= (1.2 + 2.3 + 3.4 +...+ 98.99) + (1 + 2 + 3 +...+ 98)
= [ 1.2.3 + 2.3.(4-1) +...+ 98.99.(100-97)] + [ 1.2 + 2.(3-1) + 3.(4-2) +... 98.(99-97)]
= [ 1.2.3 + 2.3.(4-1) - 1.2.3 + 3.4.(5-2) - 2.3.(4-1) +...+ 98.99.(100-97) - 97.98(99-96)] + [ 1.2 + 2.(3-1) - 1.2 + 3.(4-2) - 2.(3-1) +...+ 98.(99-97) - 97(98-96)]
= 98.99.100:3 + 98.99:2 = 323 400 + 4581 = 328251
b) B = 1.2.3 + 2.3.4 + 3.4.5 +...+ 48.49.50
4B = 1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2) +...+ 48.49.50.(51-47)
4B-B = 1.2.3.4 + 2.3.4.(5-1) - 1.2.3.4 + 3.4.5.(6-2) - 2.3.4.(5-1) +...+ 48.49.50.(51-47) - 47.48.49.(50-46)
= 48.49.50.51:4 = 1499400
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{17\cdot18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{17\cdot18}-\dfrac{1}{18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{342}\right)=\dfrac{1}{2}\cdot\dfrac{85}{171}=\dfrac{85}{342}\)
giúp mình vs ạ
a) \(A=1\cdot2\cdot3-2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\)
\(\Rightarrow4A=4\cdot\left(1\cdot2\cdot3+2\cdot3\cdot4+...+98\cdot99\cdot100\right)\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot4+3\cdot4\cdot5\cdot4+...+98\cdot99\cdot100\cdot4\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+....+98\cdot99\cdot100\cdot\left(101-97\right)\)
\(\Rightarrow4A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+3\cdot4\cdot5\cdot6-....-97\cdot98\cdot99\cdot100\)
\(\Rightarrow4A=\left(1\cdot2\cdot3\cdot4-1\cdot2\cdot3\cdot4\right)+\left(2\cdot3\cdot4\cdot5-2\cdot3\cdot4\cdot5\right)+...+98\cdot99\cdot100\cdot101\)
\(\Rightarrow4A=0+0+0+...+98\cdot99\cdot100\cdot101\)
\(\Rightarrow4A=98\cdot99\cdot100\cdot101\)
\(\Rightarrow A=\dfrac{98\cdot99\cdot100\cdot101}{4}\)