bằng shit trâu

\(4E=2^2.E=2^2+2^4+2^6+...+2^{2020}\)

\(3E=4E-E=2^{2020}-1\Rightarrow E=\frac{2^{2020}-1}{3}\)

\(3F=3^2+3^3+3^4+...+3^{101}\)

\(2F=3F-F=3^{101}-3\Rightarrow F=\frac{3^{101}-3}{2}\)

Ta có \(S=1+2+2^2+2^3+....+2^{2018}\)

Suy ra \(2S=2+2^2+2^3+2^4+....+2^{2019}\)

Nên \(2S-S=2^{2019}-1\Rightarrow S=2^{2019}-1\)

Ta có \(2^{2019}-1=2^{2016}.2^3-1=\left(2^4\right)^{504}.8-1=16^{504}.8-1\)

Vì 16 tận cùng là 6 nên \(16^{504}\)tận cùng là 6 nên \(16^{504}.8\)tận cùng là 8

Suy ra \(16^{504}.8-1\)tận cùng là 7 hay S tận cùng là 7

Vậy S =\(2^{2019}-1\)và S tận cùng là  7

a)bạn nhân lũy thừa 3 lên là tính đc, bài c thì tương tự

còn bài b mk ko bt

bạn làm ra đc ko

mk làm mẫu 1 bài nha

a) 3A=3.(1+3+3^2+.....+3^2018)

3A=3+3^2+.....+3^2019

3A-A=(3+3^2+.....+3^2019)-(1+3+3^2+.....+3^2018)

2A=3^2019-1

A=3^2019-1:2

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\(C=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2017+1\)

\(=\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)2018-\left(2018^{2019}+2018^{2018}+...+2018\right)-1\)

\(=\left(2018^{2020}+2018^{2019}+...+2018^3+2018^2\right)-\left(2018^{2019}+2018^{2018}+...+2018^2+2018\right)+1\)\(=2018^{2020}-2018+1\)

\(=2018^{2020}-2017\)

a/ Ta có:

\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)

\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)

a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)

b/ \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow B=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2017}}-\dfrac{1}{\sqrt{2018}}=1-\dfrac{1}{\sqrt{2018}}\)

\(B=2+2^2+2^3+...+2^{2016}\)

\(2B=2^2+2^3+...+2^{2017}\)

\(B=2^{2017}-2\)

các ý khác tương tự 

ý C nhân vs 3

   D                4

  E                5

3C = 3(1+3+3^2+.......+3^2017)

= 3+3^2+3^3+......+3^2018

3C - C = (3+3^2+3^3+......+3^2018) - (1+3+3^2+......+3^2017)

= 3^2018 - 1

=> C = (3^2018 - 1) : 2

còn lại tự làm nhé