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11 tháng 3 2021

\(S_2=3-3^2+3^3-3^4+...-3^{2020}\)

\(3S_2=3^2-3^3+3^4-3^5+...-3^{2021}\)

\(3S_2+S_2=3^2-3^3+3^4-3^5+...-3^{2021}+3-3^2+3^3-3^4+...-3^{2020}\)

\(4S_2=\left(3^2-3^2\right)+\left(3^3-3^3\right)+\left(3^4-3^4\right)+...+\left(3^{2020}-3^{2020}\right)+3-3^{2021}\)

\(4S_2=3-3^{2021}\)

\(S_2=\frac{3-3^{2021}}{4}\)

Xem mình đúng không?

a: =5-78*32

=5-2496

=-2491

b: \(=6\left(9-6\right)=6\cdot3=18\)

c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)

d: \(=181+3-84+8\cdot25\)

=100+200

=300

e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)

=14000-1

=13999

f: \(=3^3+25\cdot8-1=26+200=226\)

g: \(=3+2^4+1=16+4=20\)

h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)

1 tháng 10 2023

Đặt \(A=1+3+3^2+3^3+3^4+...+3^{2020}\)

\(3\cdot A=3+3^2+3^3+3^4+3^5+...+3^{2020}+3^{2021}\)

\(3A-A=3+3^2+3^3+3^4+3^5+...+3^{2020}+3^{2021}-\left(1+3+3^2+3^3+3^4+...+3^{2020}\right)\)

\(2A=3^{2021}-1\)

\(\Rightarrow A=\dfrac{3^{2021}-1}{2}\)

#\(Toru\)

1 tháng 10 2023

mình ví dụ nhá.

 

a: 6S=6+6^2+...+6^65

=>5S=6^65-1

=>S=(6^65-1)/5

b: 4S=4+4^2+...+4^401

=>3S=4^101-1

=>S=(4^101-1)/3

c: 9S=3^2+3^4+...+3^104

=>8S=3^104-1

=>S=(3^104-1)/8

27 tháng 8 2023

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

27 tháng 8 2023

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

26 tháng 9 2019

a, \(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(\Leftrightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{2022}\)

\(\Leftrightarrow3^2S-S=3^{2022}-3^0\)

\(\Leftrightarrow9S-S=3^{2022}-1\)

\(\Leftrightarrow8S=3^{2022}-1\Leftrightarrow S=\frac{3^{2022}-1}{8}\)

b,\(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)\)

\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)\)

\(=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)\)

\(=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7\)

=> đpcm

26 tháng 9 2019

Tham khảo :

a, S=30+32+34+36+...+32020

⇔32S=32+34+36+38+...+32022

⇔32S−S=32022−30

⇔9S−S=32022−1

⇔8S=32022−1⇔S=32022−18

b,S=30+32+34+36+...+32020

=(30+32+34)+(36+38+310)+...+(32016+32018+32020)

=(1+32+34)+36(1+32+34)+...+32016(1+32+34)

=(1+32+34)(1+36+...+32016)

=91(1+36+...+32016)=13.7(1+36+...+32016)⋮7 (

=> (đpcm)

=>99

10 tháng 10 2020

các pạn ơi mình cần gấp lắm lun 

giải hộ mk với

10 tháng 10 2020

Ta có: 

\(B=3+3^2+3^3+...+3^{2020}\)

\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2019}+3^{2020}\right)\)

\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)

\(B=3\cdot4+3^3\cdot4+...+3^{2019}\cdot4\)

\(B=4\cdot\left(3+3^3+...+3^{2019}\right)\) chia hết cho 4

=> đpcm