A= 1 +\(\frac{1}{3}\)+\(\frac{1}{6}\)+ .....+ \(\frac{1}{171}\)+\(\frac{1}{190}\)

A= 1 +2.(\(\frac{1}{6}\)+\(\frac{1}{12}\)+....+\(\frac{1}{342}\)+\(\frac{1}{380}\))

A=1+ 2.(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+....+\(\frac{1}{18.19}\)+\(\frac{1}{19.20}\))

A=1+2.(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+......+\(\frac{1}{18}\)-\(\frac{1}{19}\)+\(\frac{1}{19}\)-\(\frac{1}{20}\))

A=1 +2.(\(\frac{1}{2}\)-\(\frac{1}{20}\))

A=1+2.\(\frac{9}{20}\)=1+\(\frac{9}{10}\)=\(\frac{19}{10}\)

B=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+....+\(\frac{1}{2^{20}}\)

2B= 1 +\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+......+\(\frac{1}{2^{21}}\)

2B-B= 1-\(\frac{1}{2^{21}}\)

B=1-\(\frac{1}{2^{21}}\)

\(A=1+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{380}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{19.20}\)

\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=1+2\times\frac{9}{20}\)

\(=1+\frac{9}{10}\)

\(=\frac{19}{10}\)

b)\(2S=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)

\(2S=1+\frac{1}{2}+...+\frac{1}{2^{19}}\)

\(2S-S=\left(1+\frac{1}{2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)

\(S=1-\frac{1}{2^{20}}\)

c)đặt A=1+2+2^2+2^3+...+2^2006+2^2007.

2A=2(1+2+2^2+2^3+...+2^2006+2^2007)

2A=2+2^2+2^3+...+2^2008

2A-A=(2+2^2+2^3+...+2^2008)-(1+2+2^2+2^3+...+2^2006+2^2007)

A=2^2008-1

a: A=1-2-3+4+5-6-7+8+...+2001-2002-2003+2004

=(1-2-3+4)+(5-6-7+8)+...+(2001-2002-2003+2004)

=0+0+...+0

=0

b: B=1+2-3-4+5+6-7-8+...+2001+2002-2003-2004+2005+2006

=(1+2-3-4)+(5+6-7-8)+...+(2001+2002-2003-2004)+(2005+2006)

=(-4)+(-4)+...+(-4)+4011

\(=\left(-4\right)\cdot\frac{2004}{4}+4011=4011-2004=2007\)

\(a_{n-1}=\frac{1}{1+2+3+...+n}=\frac{2}{n\left(n+1\right)}\)=>\(1-a_{n-1}=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)

\(A=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)........\left(1-\frac{2}{2006.2007}\right)\)

\(=\left(\frac{1.4}{2.3}\right)\left(\frac{2.5}{3.4}\right)\left(\frac{3.6}{4.5}\right)........\left(\frac{2005.2008}{2006.2007}\right)\)\(=\frac{\left(1.2.3......2005\right)\left(4.5.6.....2008\right)}{\left(2.3.4.....2006\right)\left(3.4.5....2007\right)}=\frac{1.2008}{2006.3}=\frac{1004}{3009}\)

a) bằng 9 nha bạn

b) thì mik ko bik làm.

Đúng thì bạn tim giúp mik nha bạn. Thx bạn

Số số hạng trong dãy số 3;5;7;...;99 là:

(99-3):2+1=96:2+1=49(số)

Ta có: \(A=\left(\frac12-\frac13\right)\left(\frac12-\frac15\right)\cdot\ldots\cdot\left(\frac12-\frac{1}{99}\right)\)

\(=\frac{3-2}{2\cdot3}\cdot\frac{5-2}{2\cdot5}\cdot\ldots\cdot\frac{99-2}{2\cdot99}\)

\(=\frac{1}{2\cdot3}\cdot\frac{3}{2\cdot5}\cdot\ldots\cdot\frac{97}{2\cdot99}=\frac{1}{2^{49}\cdot99}\)

Ta có: \(A=\left(\frac12-\frac13\right)\left(\frac12-\frac15\right)\cdot\ldots\cdot\left(\frac12-\frac{1}{99}\right)\)

\(=\frac{3-2}{2\cdot3}\cdot\frac{5-2}{2\cdot5}\cdot\ldots\cdot\frac{99-2}{2\cdot99}\)

\(=\frac{1}{2\cdot3}\cdot\frac{3}{2\cdot5}\cdot\ldots\cdot\frac{97}{2\cdot99}=\frac{1}{2^{49}\cdot99}\)

a: =6/36+1/3=1/6+1/3=1/6+2/6=3/6=1/2

b: =3/4-1/2=3/4-2/4=1/4

\(a,\dfrac{3}{4}\times\dfrac{2}{9}+\dfrac{1}{3}=\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\\ b,\dfrac{1}{4}:\dfrac{1}{3}-\dfrac{1}{2}=\dfrac{1}{4}\times\dfrac{3}{1}-\dfrac{2}{4}=\dfrac{3}{4}-\dfrac{2}{4}=\dfrac{1}{4}\)