1+2+3+4+...+2016+2017+2018
1+2+3+4+...+2016+2017+2018
+) Gọi A là tổng của dãy số: 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018.
+) Số số hạng của A là:
A = (2018 - 1) : 1 + 1 = 2018.
+) Tổng A là: (2018 + 1). 2018 : 1 = 4074342.
Vậy, A = 4074342 (hay 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018 = 4074342).
+) Gọi A là tổng của dãy số: 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018.
+) Số số hạng của A là:
A = (2018 - 1) : 1 + 1 = 2018.
+) Tổng A là: (2018 + 1). 2018 : 2= 2037171
Vậy, A = 4074342 (hay 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018 = 2037171).
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
cho A =1+2^2018+3^2017+4^2016+...+2018^2+2019,B=1+2^2017+3^2016+...+2017^2+2018,chứng tỏ giá trị biểu thức A-3B dương
cho A =1+2^2018+3^2017+4^2016+...+2018^2+2019,B=1+2^2017+3^2016+...+2017^2+2018,chứng tỏ giá trị biểu thức A-3B dương
Tính tỉ số A/B biết:
A=1/2 + 1/3 + 1/4 + ... + 1/2017 + 1/2018 + 1/2019
B=2018/1 + 2017/2 + 2016/3 + ... + 2/2017 + 1/2018
\( S =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1} {2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right) \)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(\(\Rightarrow S=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\) \(\Rightarrow S=P\)\)
\(B=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(B=1+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{1}{2018}+1\right)\)
\(B=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(B=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
ta có \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
bài) Tính
a) 75%+1,2-2+1/5+20180
b) (-4/3+0,75) :2017/2018+(1+1/3-75%) :2017/2018
c) (2018-1/3-2/4-3/5-4/6-...-2018/2020) : (1/15+1/20+1/25+1/30+...1/10100)
bài) Tính
a) 75%+1,2-2+1/5+20180
b) (-4/3+0,75) :2017/2018+(1+1/3-75%) :2017/2018
c) (2018-1/3-2/4-3/5-4/6-...-2018/2020) : (1/15+1/20+1/25+1/30+...1/10100)
Giải:
a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\)
=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\)
=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\)
=\(\dfrac{7}{5}+\dfrac{-1}{4}\)
=\(\dfrac{23}{20}\)
b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\)
=\(\left[0+0\right]:\dfrac{2017}{2018}\)
=0\(:\dfrac{2017}{2018}\)
=0
c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10
1×2×3×4×...×2018×2019-1×2×3×...×2018-1×2×3×4×..×2017×20182
1x2x3x...2018x2019 - 1x2x3x..2018 - 1x2x3x4x...x2017x20182
= 1x2x3x...x2018x(2019 - 1 - 2018)
= 1x2x3x...x2018x0
= 0
so sánh 2 số A và B nếu
\(A=-\frac{1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4};B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)