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

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

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

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

\(B=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

   \(=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

    \(=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

    \(=9.\left(\frac{1}{3}-\frac{1}{675}\right)\)

    \(=9.\frac{224}{675}\)

    \(=\frac{224}{75}\)

THIẾU

\(=\frac{224}{75}< \frac{225}{75}=3\)

Vậy \(B< 3\)

mấy bọn nguuuuuuuuuuuuuuuuuu

2 h 30 chiều nay mình phả đi học rồi , làm ơn giúp mình vói mình xin cảm ơn và sẽ k

\(a,\frac{3}{20}+\frac{3}{20}+\frac{3}{42}+...+\frac{3}{132}\)

\(=3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\right)\)

\(=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\right)\)

\(=3\left(\frac{1}{4}-\frac{1}{12}\right)\)

\(=3\cdot\frac{1}{6}=\frac{1}{2}\)

a) \(\frac{3}{20}+\frac{3}{30}+\frac{3}{42}+...+\frac{3}{132}\)

= \(3\times\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\right)\)

= \(3\times\left(\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+...+\frac{1}{11\times12}\right)\)

= \(3\times\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\right)\)

= \(3\times\left(\frac{1}{4}-\frac{1}{12}\right)\)

= \(3\times\frac{1}{6}\)

= \(\frac{1}{2}\)

(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

36 x 2 + 36 x 5 + 3 x 36

= 36 x ( 2 + 3 + 5 )

= 36 x 10

= 360

=36×(2+5+3)

=36×10

=360

Do you remember me?

B =  9 . [ 4/1.3.5+4/3.5.7+4/5.7.9+...+4/25.27.29]

B =  9 . [ 1/3-1/783]

   =  9 . [ 1/3-1/783] 

   =  9 . 260/783=260/87<261/87<3

Bạn sai rồi 261/87=3 mà