Đặt :

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+................+\dfrac{9999}{10000}\)

\(A=\dfrac{1.3}{2^2}+\dfrac{2.4}{3^2}+\dfrac{3.5}{4^2}+....................+\dfrac{99.101}{100^2}\)

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+..................+\dfrac{100^2-1}{100^2}\)

\(A=\dfrac{2^2}{2^2}-\dfrac{1}{2^2}+\dfrac{3^3}{3^2}-\dfrac{1}{3^2}+............+\dfrac{100^2}{100^2}-\dfrac{1}{100^2}\)

\(A=\left(\dfrac{2^2}{2^2}+\dfrac{3^3}{3^3}+...........+\dfrac{100^2}{100^2}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^3}+........+\dfrac{1}{100^2}\right)\)

\(A=\left(1+1+........+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^3}+............+\dfrac{1}{100^2}\right)\)

\(A=99-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..........+\dfrac{1}{100^2}\right)\)

Ta có :

\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+............+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)\(\dfrac{1}{2^2}+........+\dfrac{1}{100^2}< \dfrac{1}{1}-\dfrac{1}{2}+.......+\dfrac{1}{99}-\dfrac{1}{100}\)\(\Rightarrow\dfrac{1}{2^2}+.........+\dfrac{1}{100^2}< 1-\dfrac{1}{100}\)

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.........+\dfrac{1}{100^2}< \dfrac{100}{101}\)

\(\Rightarrow99-\left(\dfrac{1}{2^2}+...........+\dfrac{1}{100^2}\right)< 99-\dfrac{100}{101}\)

\(\Rightarrow A< 99-\dfrac{100}{101}\)

\(\Rightarrow a< 99\rightarrowđpcm\)

~ Học tốt ~

3/4.8/9.15/16...9999/10000

=\(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\)...\(\dfrac{99.101}{100.100}\)

=\(\dfrac{1.2...99}{2.3.100}\).\(\dfrac{3.4...101}{2.3.100}\)

=\(\dfrac{1}{100}\).\(\dfrac{101}{2}\)

=\(\dfrac{101}{200}\)

\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}\)+...+\(\frac{9999}{10000}\)

= (1-\(\frac{1}{4}\)) +(1-\(\frac{1}{9}\))+(1-\(\frac{1}{16}\))+...+(1-\(\frac{1}{10000}\))

= 99 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)) => 99 - A

 Dễ thấy A>0 =>S < 99 (1)

Lại có A= \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\)

=> A<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)

=>A<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...\(\frac{1}{99}\)-\(\frac{1}{100}\)

=>A<1-\(\frac{1}{100}\)<1

...

Hzzz tự làm ko nổi

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

\(=\frac{\left(1.2.3....99\right)\left(3.4.5....101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)

\(=\frac{1.101}{100.2}=\frac{101}{200}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{99.101}{100^2}\)

\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)

= 3 . 8 . 15 .... 9999 / 4 . 9 . 16 .... 10000

= ( 1 . 3 ) . ( 2 . 4 ) .( 3 . 5) .... ( 99 .... 101 ) / ( 2. 2) . (3.3). (4.4)...(100.100)

= 1. 101/100.2

= 101/ 200

k nha , đúng đó

1*3/2*2.2*4/3*3.3*5/4*4.....99*101/100*100.                                                                                                                                          =1*2*3*...*99/2*3*4*...*100.3*4*5*...*101/2*3*4*...*100.                                                                                                                          =1/100 . 101/2.                                                                                                                                                                               =101/200.

   Tính tổng:  \(A=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+...+\frac{9999}{100^2}\)

Giúp mình nha. Ai nhanh mình sẽ k cho!

3/4 . 8/9 . 15/16 ... 9999/10000

= 1.3/2.2 . 2.4/3.3 ... 99.101/100.100

= 1 . 2 . ... . 99 / 2 . 3 . 100 × 3 . 4 ... 101 / 2 . 3 ... 100

= 1 / 100 . 101 / 2

= 101 / 200

=1.3/2.2 .2.4/3.3 .3.5/4.4 . ...... 99.101/100.100

=1.2.3.4.5 ......  .99/2.3.4.....100   .    3.4.5 ....... .101/2.3.4.5   ....  .100

=1/100 .101/2

=101/200

k cho mink nha

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)

\(=\frac{3.8.15...9999}{4.9.16...10000}\)

\(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(99.101\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(100.100\right)}\)

\(=\frac{\left(1.2.3...99\right).\left(3.4.5...101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}\)

\(=\frac{1.101}{100.2}\)

\(=\frac{101}{200}\)