Tính A= 1+(1+2)+(1+2+3)+........+(1+2+3+.....+2017)/1.2+2.3+3.4+.......+2017.2018
Tính A=1+(1+2)+(1+2+3)+....+(1+2+3+.....+2017)/1.2+2.3+3.4+......+2017.2018
S = 1/1.2 + 1/2.3 + 1/3.4 +..........1/2017.2018 + 1/2018.2019
can gap
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\)
b)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
c)\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
tính các tổng sau:
A=1.2+2.3+3.4+...+n(n+1)
B=1.2.3+2.3.4+...+n(n+1)(n+2)
C=1.2+3.4+5.6+...+2017.2018
D=1.4+2.5+3.6+...+n(n+3)
Giúp mk nha, ai nhanh mk k!
a,1/1.2 + 1/2.3 + 1/ 3.4+....+ 1/2017.2018
b, 2/3.5+ 2/ 5.7 + 2/7.9 +....+ 2/.91.99
Ai giúp mink tích cho!
Tính A = 1.2^2+2.3^2+3.4^2+....+2017.2018^2
Tính
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{3}+\frac{1}{5}+\frac{1}{35}+...+\frac{1}{9999}\)
Tính tổng các ps sau
a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
b,\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2015.2017}\)
Tính tổng sau một cách hợp lí :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
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