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Đặt A=(đã cho).
=>2A=2/1*2*3+2/2*3*4+2/3*4*5+...+2/37*38*39.
=>2A=1/1*2-1/2*3+1/2*3-1/3*4+...+1/37*38-1/38*39.
=>2A=1/1*2=1/38*39.
Đến đây tự bấm máy nha.
tk mk nha.
chắc chắn đúng,nay mk làm bài này.
-chúc ai tk mk học giỏi-
= 1/2.(2/1.2.3+2/2.3.4+.....+2/50.51.52
=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+....+1/50.51-1/51.52
=1/2.(1/1.2-1/51.52)
=1/2.(1/2-1/2652)
=1/2.1325/2652
=1325/5304
A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/50.51-1/51.52
A=1/1.2-1/51.52
phần còn lại tự giải nhé
Ta có công thức:
\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)
vậy
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)
mình làm đầu tiên đó,
Chúc bạn học tốt !
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)
\(=\frac{1}{2}\cdot\frac{77}{156}\)
\(=\frac{77}{312}\)
Trả lời
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{18\cdot19\cdot20}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}\)
\(=\frac{19}{20}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\left(\frac{190}{380}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}\)
\(=\frac{189}{760}\)
Chúc bạn học tốt !!!
a ) Co :
1/1.2 - 1/2.3 = 2/1.2.3
1/2.3 - 1/3.4 = 2/2.3.4
...
1/37.38 - 1/38.39 = 2/37.38.39
=> 2M = 2/1.2.3 + 2/2.3.4 + ... + 2/37.38.39
=> 2M = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/37.38 - 1/38.39
=> 2M = 1/2 - 1/1482
=> 2M = 370/741
=> M = 185/741
B ) A = 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8
3A = 1 + 1/3 + 1/3^2 + ... + 1/3^7
3A - A = ( 1 + 1/3 + 1/3^2 + ... + 1/3^7 ) - ( 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^8 )
2A = 1 - 1/3^8
A = ( 1 - 1/3^8 ) / 2
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\left(\frac{4950-1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
Ủng hộ mk nha!!
2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 1/18.19.20
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 +...+1/18.19 - 1/19.20
2A = 1/1.2 - 1/19.20
2A = 1/2 - 1/19.20
A = (1/2 - 1/19.20) : 2
A = 1/4 - 1/(19.20.2)
MÀ 1/(19.20.2) > 0
nên A<1/4
Đặt A=\(\frac{-1}{2\cdot3\cdot4}\)+\(\frac{-1}{3\cdot4\cdot5}\)+...+\(\frac{-1}{12\cdot13\cdot14}\)
A=(-1)*(\(\frac{1}{2\cdot3\cdot4}\)+\(\frac{1}{3\cdot4\cdot5}\)+...+\(\frac{1}{12\cdot13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{12\cdot13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{12\cdot13}\)-\(\frac{1}{13\cdot14}\))
A=\(\frac{-1}{2}\)*(\(\frac{1}{6}\)-\(\frac{1}{182}\))
A=\(\frac{-1}{2}\)*\(\frac{44}{273}\)
A=\(\frac{-22}{273}\)
\(\frac{-1}{2.3.4}+\frac{-1}{3.4.5}+\frac{-1}{4.5.6}+...+\frac{-1}{11.12.13}+\frac{-1}{12.13.14}\)
\(=-\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+...+\frac{2}{11.12.13}+\frac{2}{12.13.14}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{11.12}-\frac{1}{12.13}+\frac{1}{12.13}-\frac{1}{13.14}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{13.14}\right)=-\frac{1}{2}.\frac{44}{273}=-\frac{22}{273}\)