(1+2+3+...+2017+2018+2017+...+3+2+1) / 2018
1+2018+2*2017+3*2016+..........................+2016*3+2017+2+2018*1
1+(1+2)+(1+2+3)+..........................(1+2+3+...............+2017+2018
A= 1+(1+2) + (1+2+3) +...+( 1+2+3+...+2018) - (1×2018+2×2017+...+2017×2+2018×1)
Giúp mình với
Cho tổng A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+\frac{2018}{2017^2+3}+...+\frac{2018}{2017^2+n}+...+\frac{2018}{2017^2+2017}\)
(A có 2017 số hạng). Chứng tỏ A không là số nguyên
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
45612223698++56456+89575637259415767549846574257
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 \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}=\frac{1}{2019}\)
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).
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 :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)]