(36 - 1²).(36 - 2²).(36 - 3²)...(36 - 20²)

= (36 - 1²).(36 - 2²)...(36 - 6²)...(36 - 20²)

= (36 - 1²).(36 - 2²)...(36 - 36)...(36 - 20²)

= (36 - 1²)(36 - 2²)...0...(36 - 20²)

= 0

E = \(\frac{36}{1\cdot7}+\frac{36}{7\cdot13}+...+\frac{36}{94\cdot100}=\frac{36}{6}\left[\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+...+\frac{1}{94\cdot100}\right]\)

\(=6\left[1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{94}-\frac{1}{100}\right]=6\left[1-\frac{1}{100}\right]\)

\(=6\cdot\frac{99}{100}=\frac{297}{50}\)

F = \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{\left[3a+2\right]\left[3a+5\right]}\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3a+2}-\frac{1}{3a+5}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3a+5}\right]=\frac{1}{6}-\frac{1}{9a+15}\)

G = \(\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{4}{8\cdot12}+\frac{5}{12\cdot17}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{12}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

E=36/1-36/7+36/7-36/13+...+36/94-36/100

  =36-36/100=891/25

Ta có: \(\frac{1}{\left(2n+1\right)\left(2n+3\right)}-\frac{1}{\left(2n+3\right)\left(2n+5\right)}\)

\(=\frac{2n+5-\left(2n+1\right)}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}\)

\(=\frac{4}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}\)

=>\(\frac{1}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}=\frac14\left(\frac{1}{\left(2n+1\right)\left(2n+3\right)}-\frac{1}{\left(2n+3\right)\left(2n+5\right)}\right)\)

Do đó, ta có:

\(\frac{1}{1\cdot3\cdot5}=\frac14\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}\right)\)

\(\frac{1}{3\cdot5\cdot7}=\frac14\left(\frac{1}{3\cdot5}-\frac{1}{5\cdot7}\right)\)

...

\(\frac{1}{25\cdot27\cdot29}=\frac14\left(\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right)\)

Do đó: \(\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\cdots+\frac{1}{25\cdot27\cdot29}=\frac14\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+\cdots+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right)\)

\(=\frac14\left(\frac{1}{1\cdot3}-\frac{1}{27\cdot29}\right)\)

=>\(36\left(\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\cdots+\frac{1}{25\cdot27\cdot29}\right)=36\cdot\frac14\left(\frac{1}{1\cdot3}-\frac{1}{27\cdot29}\right)\)

=>\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\cdots+\frac{36}{25\cdot27\cdot29}=9\left(\frac13-\frac{1}{27\cdot29}\right)=3-\frac{1}{3\cdot29}=\frac{3\cdot3\cdot29-1}{3\cdot29}=\frac{260}{87}\)

a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760

2A=\(\frac{2}{1.2.3}\)+\(\frac{2}{2.3.4}\)+...+\(\frac{2}{18.19.20}\)

=1/1.2-1/2.3+1/2.3-1/3.4+...+1/18.19-1/19.20

=1/2-1/19.20

A=1/4-1/19.20.2

vậy A<1/4

a) 26 x 11 = 26 x (10 + 1)

= 26 x 10 + 26 x 1 = 260 + 26 = 286

35 x 101 = 35 x (100 + 1)

= 35 x 100 + 35 x 1 = 3500 + 35 = 3535

b) 213 x 11 = 213 x (10 +1)

= 213 x 100 + 213 x 1 = 2130 + 213 = 2343

123 x 101 = 123 x (100 + 1)

= 123 x 100 + 123 x 1

= 12300 + 123 = 12423

Nói thêm: Muốn nhân một số với 11 (hoặc 101) ta nhân số đó với 10 (hoặc 100) rồi cộng thêm chính số đó.

$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$

D

D