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IL
0
CU
11 tháng 2 2019
A = (-1)(-1)^2(-1)^3...(-1)^2019
A = (-1)^1+2+3+...+2019
A = (-1)^2039190
A = 1
S = 1.2.3 + 2.3.4 + 3.4.5 + ... + 2018.2019.2020
4S = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + .... + 2018.2019.2020.4
4S = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 2018.2019.2020.(2021 - 2017)
4S = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2018.2019.2020.2021 - 2017.2018.2019
4S = 2018.2019.2020.2021
S = 2018.2019.2020.2021 : 4 = ...
NT
2
XO
18 tháng 2 2020
\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)
\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)
\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)
Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
21 tháng 2 2021
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TN
0
NN
2
DP
18 tháng 5 2018
\(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2018}\)
\(\frac{3}{2}A=\frac{3}{2}\cdot\left[\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2018}\right]+\frac{1}{2}\)
\(\frac{3}{2}A=\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2019}+\frac{1}{2}\)
\(\frac{3}{2}A=\left[\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2019}\right]-\left[\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2018}\right]+\frac{1}{2}\)
\(\frac{3}{2}A=\left(\frac{3}{2}\right)^{2019}-\frac{3}{2}+\frac{1}{2}\)
Còn lại bn tự làm nốt
TD
0
TT
1
27 tháng 9 2020
\(B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{2018}.\frac{\left(1+2018\right).2018}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{2019}{2}=1+\frac{3+4+...+2019}{2}=1+\frac{\left(3+2019\right)2017}{2}=2039188\)
\(B=3+3^2+3^3+...+3^{2018}\)
\(3B=3.\left(3+3^2+3^3+...+3^{2018}\right)\)
\(3B=3^2+3^3+3^4+...+3^{2019}\)
\(3B-B=\left(3^2+3^3+3^4+...+3^{2019}\right)-\left(3+3^2+3^3+...+3^{2018}\right)\)
\(2B=3^{2019}-3\)
\(\Rightarrow B=\dfrac{3^{2019}-3}{2}\)
\(#WendyDang\)
\(B=3^1+3^2+3^3+...+3^{2018}\)
\(3\cdot B=3^2+3^3+3^4+...+3^{2019}\)
\(B=(3^{2019}-3):2\)