1 + 1/2 . (1+2) + 1/3 . (1 + 2 + 3) +...+ 1/20 . (1 + 2 + 3 +...+ 20)
= 1 + 1/2 . 2 . 3 : 2 + 1/3 . 3 . 4 : 2 + ... + 1/20 . 20 . 21 : 2
= 2/2 + 3/2 + ... + 21/2
= 2 + 3 + ... + 21 / 2
= 230/2 = 115.
1 + 1/2 . (1+2) + 1/3 . (1 + 2 + 3) +...+ 1/20 . (1 + 2 + 3 +...+ 20)
= 1 + 1/2 . 2 . 3 : 2 + 1/3 . 3 . 4 : 2 + ... + 1/20 . 20 . 21 : 2
= 2/2 + 3/2 + ... + 21/2
= 2 + 3 + ... + 21 / 2
= 230/2 = 115.
Tính :\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}}{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}\)
Giải rõ ra giùm mik nha! ~~ Thank you~~
(1-\(\frac{1}{1+2}\)) . (1-\(\frac{1}{1+2+3}\)).(1-\(\frac{1}{1+2+3+4}\))*...*(1-\(\frac{1}{1+2+...+2016}\))
GIẢI BÀI NÀY GIÙM MIK NHA! :))))))))))))
Tính A=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{20}\left(1+2+3+...+20\right)\)
Ai giúp mình với
1 +\(\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}+\left(1+2+3+4\right)+...+\frac{1}{20}\left(1+2+3+4+.....+20\right)\)
giúp mình với ai nhanh mình tick cho
mấy bạn giải giúp mk với B=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+..+20\right)\)
1. a) Chứng minh: \(\frac{1}{6}<\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+........+\frac{1}{100^2}<\frac{1}{4}\)
b) Tìm số nguyên a để: \(\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}\)là số nguyên
2. Tính: \(A=\frac{10\frac{1}{3}\left(26\frac{1}{3}-\frac{176}{7}\right)-\frac{12}{11}\left(\frac{10}{3}-1,75\right)}{\frac{5}{91-0,25.\frac{60}{11}-1}}\)
3. Biết rằng: \(1^2+2^2+3^2+...+10^2=385\)
Tính tổng \(S=2^2+4^2+...+20^2\)
Giúp mik nha mik đg rất gấp
Ai làm nhanh nhất mik sẽ tick
Tính \(B=
1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+2+...+20\right)\)
Giúp mình nhé !
B = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+....+\frac{1}{20}\left(1+2+3+...+20\right)\)
Tính A = \(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+...+\frac{2+3+4+...+20}{1+2+3+4+...+20}\)