\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\left(\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)=\frac{1}{2}.\left(1-\frac{1}{2011}\right)=\frac{1}{2}.\frac{2010}{2011}=\frac{1005}{2011}\)

1/.3 + 1/3.5 + 1/5.7 + ... + 1/2009.2011

= 1/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2009.2011)

= 1/2 . (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2009 - 1/2011)

= 1/2 . (1 - 1/2011)

= 1/2 . 2010/2011

= 1005/2011

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2009.2011}\)

\(=\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}x\left(1-\frac{1}{2011}\right)\)

\(=\frac{1005}{2011}\)

Program HOC24;

var b: real;

i,n: integer;

begin

write('Nhap n='); readln(n);

b:=0;

for i:=1 to n do b:=b+1/(i+2);

write('B= ',b:1:2);

readln

end.

thanh niên điêu

a, 1 + 2 + 3 + ... + x = 120

=> (x+1)x/2 = 120

=>x(x +1)=120.2=240

=>15.16 = 240

=>x=15

Vậy x=15

Phần b làm tương tự

c, x - ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/53.55) = 3/5

=> x = 3/5 + ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/53.55)

=> x = 3/5 + ( 1-1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/53 - 1/55 )

=> x = 3/5 + ( 1- 1/55 )

=> x = 3/5 + 54/55

=> x = 87/55

Vậy x = 87/55

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

PTHH :

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,4                    0,4         0,4 

\(b,V_{H_2}=0,4.22,4=8,96\left(l\right)\)

\(m_{FeCl_2}=127.0,4=50,8\left(g\right)\)

\(m_{ddHCl}=22,4+200-\left(0,4.2\right)=221,6\left(g\right)\)

\(C\%=\dfrac{50,8}{221,6}.100\%\approx22,92\%\)

a) Fe + 2HCl -> FeCl2 + H2

b) Ta có: n_Fe=m/M=22,4/56=4 (mol)

n_Fe=n_FeCl2=n_H2= 4 (mol)

Thể tích khí hidro thu được ở đktc: V_H2=n.22,4=4 . 22,4= 89,6 (l)

c)Nồng độ phần trăm của dd thu đc sau pứ:

C%=m_ct.100/mdd= 22,4.100/200=11,2 %

S=(1/1.3+1/3.5+.....+1/7.9)+(1/2.4+1/4.8+1/8.10)

2S=1/2.(1-1/3+1/5-1/5+....+1/7-1/9)+(1/2-1/4+1/4-1/8+1/8-1/10)

2S=1/2.(1-1/9)+(1/2-1/10)

2S=1/2.(8/9+2/5)

=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+1/9.10

=1/2(1-1/9)+1/90

=1/2.8/9+1/90=4/9+1/90=41/90