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a,S=1-3+32-33+.............+398-399
S=(1-3+32-33)+.............+(396-397+398-399)
S=(-20)+...............+396.(1-3+32-33)
S=(-20)+.............+396.(-20)
S=(1+34+...........+396).(-20) chia hết cho 20
b,3S=3-32+33-34+..............+399-3100
3S+S=(3-32+33-34+...........+399-3100)+(1-3+32-33+..............+398-399)
4S=1-3100
S=\(\frac{1-3^{100}}{4}\)
S = (1+3+3^2)+(3^3+3^4+3^5)+.....+(3^97+3^98+3^99)
= 10+3^3.(1+3+3^2)+.....+3^97.(1+3+3^2)
= 10+3^3.10+.....+3^97.10
= 10.(1+3^3+....+3^97) chia hết cho 10
Mà 10 chia hết cho 5 => S chia hết cho 5
k mk nha
a) Đặt biểu thức trên là A, ta có:
A = 21 + 22 + 23 + 24 + ... + 299 + 2100
=> A = (21 + 22) + (23 + 24) + ... + (299 + 2100)
=> A = 21.(1 + 2) + 23.(1 + 2) + ... + 299.(1 + 2)
=> A = 21.3 + 23.3 + ... + 299.3
=> A = 3(21 + 23 + ... + 299)
=> A ⋮ 3
\(26=13.2\)
\(s=3.\left(1+3+9\right)+3^4.\left(1+3+9\right)+....+3^{2012}.\left(1+3+9\right)\)
\(s=3.13+3^413+.....+3^{2012}.13\)
\(s=13.\left(3+3^4+....+3^{2012}\right)\)
\(\Rightarrow s=3.\left(1+3\right)+3^3.\left(1+3\right)+.......+3^{2015}.\left(1+3\right)\)
\(s=3.4+3^3.4+....+3^{2015}.4\)
\(s=4.\left(3+3^3+.....+3^{2015}\right)\)
\(\Rightarrow4⋮2\Rightarrow4.\left(3+3^3+....+3^{2015}\right)⋮2\)
\(\Rightarrow s⋮2\Leftrightarrow s⋮13\)
\(\Rightarrow s⋮\orbr{\begin{cases}13\\2\end{cases}}\Leftrightarrow s⋮26\)
a=(1-3+3^2-3^3)+(3^4-3^5...+(3^96-3^97+3^98-3^99)
a=(1-3+3^2-3^3)+3^4x(1-3+3^2-3^3)+...+3^96x(1-3+3^2-3^3)
a=(-20)+3^4x(-20)+...+3^96x(-20)
a=(-20)+(3^4+3^8+...+3^96)
vi-20chia het cho 4=>achia hetcho 4
Lời giải:
a)
\(A=1-3+3^2-3^3+3^4-3^5+..+3^{98}-3^{99}\)
\(=(1-3+3^2-3^3)+(3^4-3^5+3^6-3^7)+....+(3^{96}-3^{97}+3^{98}-3^{99})\)
\(=(1-3+3^2-3^3)+3^4(1-3+3^2-3^3)+...+3^{96}(1-3+3^2-3^3)\)
\(=(1-3+3^2-3^3)(1+3^4+...+3^{96})=-20(1+3^4+...+3^{96})\vdots 20\)
Vậy $A$ chia hết cho $20$
b)
\(A=1-3+3^2-3^3+3^4-3^5+...+3^{98}-3^{99}\)
\(3A=3-3^2+3^3-3^4+3^5-3^6+...+3^{99}-3^{100}\)
Cộng theo vế:
\(\Rightarrow A+3A=1-3^{100}\)
\(\Rightarrow A=\frac{1-3^{100}}{4}\)
a) \(B=1+3+3^2+3^3+....+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+2^3\right)+....+3^{96}\left(1+3+3^2+3^3\right)\)
\(=\left(1+3+3^2+3^3\right)\left(1+3^4+...+3^{96}\right)\)
\(=40\left(1+3^4+....+3^{96}\right)\)\(⋮\)\(40\)
b) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=40.3^4\)
a,S=(1-3+32-33)+............+(396-397+398-399)
S=(-20)+...................+396.(1-3+32-33)
S=(-20)+................+396.(-20)
S=(1+34+........+396).(-20) chia hết cho 20(đpcm)
b,3S=3-32+33-34+..............+399-3100
3S+S=(3-32+33-34+.............+399-3100)+(1-3+32-33+...............+398-399)
4S=-3100+1
S=\(\frac{-3^{100}+1}{4}\)