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\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
1.
B = 3100 - 399 + 398 - 397 + ... + 32 - 3 + 1
3B = 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3
3B + B = ( 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3 ) + ( 3100 - 399 + 398 - 397 + ... + 32 - 3 + 1 )
4B = 3101 + 1
B = \(\frac{3^{101}+1}{4}\)
`A=1+4+4^2+4^3+....+4^99+4^100`
`=>4A=4+4^2+4^3+4^4+...+4^100+4^101`
`=>4A-A=4^101-1`
`=>3A=4^101-1`
`=>A=(4^101-1)/3`
Ta có: \(A=1+4+4^2+...+4^{99}+4^{100}\)
\(\Leftrightarrow4\cdot A=4+4^2+4^3+...+4^{100}+4^{101}\)
\(\Leftrightarrow4\cdot A-A=4^{101}-1\)
hay \(A=\dfrac{4^{101}-1}{3}\)
tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^
\(a)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{22}{132}+\frac{11}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{33}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{34}{132}+\frac{1}{20}\)
\(=\frac{17}{66}+\frac{1}{20}\)
\(=\frac{203}{660}\)
\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\frac{3}{10}+\frac{1}{132}\)
\(=\frac{198}{660}+\frac{5}{660}\)
\(=\frac{203}{660}\)
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)....\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right)......\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{100}{99}=\frac{3.4.5.6.....100}{2.3.4.5...99}=\frac{100}{2}=50\)
Vậy T = 50
T={1/2 + 1} .{1/3+1} . {1/4+1}....{ 1/98+1].[1/99+1}
=3/2 . 4/3 . 5/4.....99/98 . 100/99
=1/2 . 1 . 1.....1. 100/1 (MIK RÚT GỌN CHÉO CÁC PHÂN SỐ LIỀN NHAU)
=100/2
=50 Nhé
\(a;\frac{1}{n}-\frac{1}{n-1}=\frac{n-1-n}{n\left(n-1\right)}=-\frac{1}{n\left(n-1\right)}\)
a) \(\frac{1}{n}-\frac{1}{n-1}=\frac{n-1-n}{n\left(n-1\right)}=-\frac{1}{n\left(n-1\right)}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)(cái này là 1 tính chất nha bn ! tìm hiểu thêm nhé )
c)đặt C= \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
=> 2C = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
=> C=5/39
d) Ý d) lm tương tự ý c nha
e) đặt E =\(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
=> 2E=\(1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
lấy 2E-E =\(1+\frac{1}{2}+...+\frac{1}{2^{99}}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{100}}=1-\frac{1}{2^{100}}\)
=.> E=1 - \(\frac{1}{2^{100}}\)
\(B=1+\frac{1}{2}+\frac{1}{2}^2+\frac{1}{2}^3+...+\frac{1}{2}^{100}\)
\(B=1+\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+...+\frac{1^{100}}{2^{100}}\)
\(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2B-B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{100}}\)
\(B=2-\frac{1}{2^{100}}=\frac{2^{99}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{99}-1}{2^{100}}\)
\(T=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(T=\frac{1}{2}.100\)
\(T=50\)
A = 1 + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +.......+\(\dfrac{1}{3^{n-1}}\) + \(\dfrac{1}{3^n}\)
3\(\times\) A = 3 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+........+ \(\dfrac{1}{3^{n-1}}\)
3A - A = 3 + \(\dfrac{1}{3}\) - 1 - \(\dfrac{1}{3^n}\)
2A = \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)
A = ( \(\dfrac{7}{3}\) - \(\dfrac{1}{3^n}\)): 2
A = \(\dfrac{7.3^{n-1}-1}{3^n}\) : 2
A = \(\dfrac{7.3^{n-1}-1}{2.3^n}\)
B = \(\dfrac{1}{2}\) - \(\dfrac{1}{2^2}\) + \(\dfrac{1}{2^3}\) - \(\dfrac{1}{2^4}\)+......+\(\dfrac{1}{2^{99}}\) - \(\dfrac{1}{2^{100}}\)
2B = 2 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2^2}\) - \(\dfrac{1}{2^3}\)+ \(\dfrac{1}{2^4}\)-.......-\(\dfrac{1}{2^{99}}\)
2B + B = 2 - \(\dfrac{1}{2^{100}}\)
3B = 2 - \(\dfrac{1}{2^{100}}\)
B = ( 2 - \(\dfrac{1}{2^{100}}\)): 3
B = \(\dfrac{2.2^{100}-1}{2^{100}}\) : 3
B = \(\dfrac{2^{101}-1}{3.2^{100}}\)
C = \(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)....\left(\frac{1}{99}+1\right)\)
\(\Rightarrow C=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
\(\Rightarrow C=\frac{3.4.5...100}{2.3.4...99}\)
\(\Rightarrow C=\frac{100}{2}=50\)
--- Dấu “ \(.\)“ là dấu nhân
=>C=3/2.4/3.5/4.....100/99=3.4.5.....100/2.3.4.....99=100/2=50.