a)99/100

Ta thay:1/6=1.6; 1/66=6.11; 1/176= 11.16; 1/336= 16.21;...........

=1/6+1/66+1/176+1/376+.....+1/496.501

=1/5.(1-1/501)

=1/5=500/501=100/501

Vay y= 100/501

\(y=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{496.501}\)

\(\Rightarrow y=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)

\(\Rightarrow5y=5.(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{496.501}\)

\(\Rightarrow5y=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{496.501}\)

\(\Rightarrow5y=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{496}-\frac{1}{501}\)

\(\Rightarrow5y=1-\frac{1}{501}\)

\(\Rightarrow5y=\frac{501}{501}-\frac{1}{501}\)

\(\Rightarrow5y=\frac{500}{501}\)

\(\Rightarrow y=\frac{500}{501}\div5\)

\(\Rightarrow y=\frac{500}{501}.\frac{1}{5}\)

\(\Rightarrow y=\frac{100}{501}\)

\(D=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+\frac{1}{546}\)

\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+\frac{1}{21.26}\)

\(D=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)

\(D=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{26}\right)\)

\(D=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{26}\right)\)

\(D=\frac{1}{5}.\frac{25}{26}=\frac{5}{26}\)

\(x+\frac{5}{3}=\frac{1}{81}\)

\(x=\frac{1}{81}-\frac{5}{3}\)

\(x=\frac{-134}{81}\)

\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+...+\frac{1}{1886}\)

\(=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{41.46}\)

\(=\frac{5}{5}\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{41.46}\right)\)

\(=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{41.46}\right)\)

\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{41}-\frac{1}{46}\right)\)

\(=\frac{1}{5}\left(1-\frac{1}{46}\right)\)

\(=\frac{1}{5}.\frac{45}{46}=\frac{9}{46}\)

1/(1.6) + 1/(6.11) + 1/(11.16) + … + 1/[(5n-4)(5n+1)] 
=(1/1 – 1/6)/5 + (1/6 – 1/11)/5 + (1/11 – 1/16)/5 +…+ [1/(5n-4) – 1/(5n+1)]/5 
=[1/1 – 1/6 + 1/6 – 1/11 + 1/11 – 1/16 + … + 1/(5n-4) – 1/(5n+1)]/5 
=[1 – 1/(5n+1)]/5 

=[1 – 1/(5.100+1)]/5 = 100/501

Quy luật: 
6 = 1.6 
66 = 6.11 
176 = 11.16 
336 = 16.21 
... 
1/(1.6) + 1/(6.11) + 1/(11.16) + … + 1/[(5n-4)(5n+1)] 
=(1/1 – 1/6)/5 + (1/6 – 1/11)/5 + (1/11 – 1/16)/5 +…+ [1/(5n-4) – 1/(5n+1)]/5 
=[1/1 – 1/6 + 1/6 – 1/11 + 1/11 – 1/16 + … + 1/(5n-4) – 1/(5n+1)]/5 
=[1 – 1/(5n+1)]/5 
Tổng 100 số đầu =[1 – 1/(5.100+1)]/5 = 100/501

1/1.6 + 1/6.11+ 1/11.16+ .... 
số thứ 100 có dạng 1/(496.501) 
do đó tổng trên bằng 1/5( 1/1- 1/501) = 100/ 501

hc tốt

\(5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{496.501}\)

\(5B=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+\frac{21-16}{16.21}+...+\frac{501-496}{496.501}\)

\(5B=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{496}-\frac{1}{501}\)

\(5B=1-\frac{1}{501}=\frac{500}{501}\Rightarrow B=\frac{100}{501}\)

*Số thứ 100 của dãy là : \(\frac{1}{100.101}\)

Ta có :

    \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(=1-\frac{1}{101}\)

\(=\frac{100}{101}\)

Bài 2 mình cho quy luật rồi bạn tự tính nhé.

Quy luật :

66 = 6 +60

176 = 66 + 110

336 = 176 +160

\(...\)

Số tiếp theo bạn cứ việc lấy số trước nó cộng với số hạng thứ hai cộng với 50 (lấy số hạng thứ hai cộng với 50) trong phép tính trước là ra.

1A  = 1/6 . (1 - 1/501) = 1/6 . 500/501 => A = 500/501.6=500/3006

=1/1*6+1/6*11+1/11*16+1/16*31+...+1/496+1/496*501

=1/5*(1-1/6*1/6-1/11+1/11-1/16+1/16-1/31+...+1/496-1/501)

=1/5*(1-1/501)

=1/5*500/501

=100/101

Vậy A=100/101

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với