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\(B=\frac{3}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{49.50.51}\right)\)
\(=\frac{3}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2550}\right)\)
\(=\frac{3}{2}\cdot\frac{637}{1275}\)
\(=\frac{637}{850}\)
bạn phải cho ra 2 số cuối thì mới làm đc nha có 1 s
ố cuối ko làm đc đâu
A= 1-1/2 + 1-1/3 + 1/2-1/5 + 1/3-1/8+ 1/5-1/13+1/8- 1/21 +....+ 1/610- 1/1597
A= 1/610
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
a) \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{1.2}-\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{9900}\)
\(A=\frac{9898}{19800}.\)
Vậy :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{9898}{19800}:2\)
\(A=\frac{4949}{19800}.\)
a) 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 = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
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 = \(\frac{1}{2}.\frac{4949}{9900}\)
A = \(\frac{4949}{19800}\)
3/1*3+3/3*5+......+3/99*101
=3/2*(2/1*3+3/3*5+.............+2/99*101)
=3/2*(1-1/3+1/3-1/5+..........+1/99-1/101)
=3/2*(1-1/101)
=3/2*100/101
=150/101
Câu 1
=>S=2/3( 2/(1.3) + 2/(3.5)+.....+ 2/(99.101) )
=>S=2/3(1-1/3+1/3-1/5+...+1/99-1/101)
=>S=2/3(1-1/101)
=>S=2/3.100/101
=>S=200/303
2/ \(\frac{2}{3}S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{25-23}{23.24.25}\)
\(\frac{2}{3}S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}=\frac{1}{2}-\frac{1}{24.25}\Rightarrow S=\left(\frac{1}{2}-\frac{1}{24.25}\right):\frac{2}{3}\)
1/
\(\frac{2}{3}S=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\)
\(\frac{2}{3}S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\Rightarrow S=\frac{100}{101}.\frac{3}{2}=\frac{150}{101}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
ta có A=3/1*3+3/3*5+3/5*7+...+3/49*51
=> A=3*1/2*(2/1*3+2/3*5+..+2/49*51)
=> A=3/2*(1-1/3+1/3-1/5+..+1/49-1/51)
=> A=3/2*(1-1/51)
=> A= 3/2* 50/51
=> A= 25/17
đề câu a sai ở tử của số hạng thứ 2
câu a phải là như z ms làm được bn ơi
A = 31.3+33.5+...+319.2031.3+13.5+...+319.20