Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\left(2+...+2^{19}\right)⋮7\)
a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{19}\right)⋮7\)
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
Tham khảo :
a, S=30+32+34+36+...+32020S=30+32+34+36+...+32020
⇔32S=32+34+36+38+...+32022⇔32S=32+34+36+38+...+32022
⇔32S−S=32022−30⇔32S−S=32022−30
⇔9S−S=32022−1⇔9S−S=32022−1
⇔8S=32022−1⇔S=32022−18⇔8S=32022−1⇔S=32022−18
b,S=30+32+34+36+...+32020S=30+32+34+36+...+32020
=(30+32+34)+(36+38+310)+...+(32016+32018+32020)=(30+32+34)+(36+38+310)+...+(32016+32018+32020)
=(1+32+34)+36(1+32+34)+...+32016(1+32+34)=(1+32+34)+36(1+32+34)+...+32016(1+32+34)
=(1+32+34)(1+36+...+32016)=(1+32+34)(1+36+...+32016)
=91(1+36+...+32016)=13.7(1+36+...+32016)⋮7=91(1+36+...+32016)=13.7(1+36+...+32016)⋮7 (
=> (đpcm)
=>99
a,S=1+3+32+...+360
3S=3+32+33+...+361
3S-S=(3+32+33+...+361)-(1+3+32+...+360)
2S = 361 - 1
b,2S+1=361-1+1=361 = 3x-3
=>x-3=61=>x=64
c, S=1+3+32+...+360
=(1+3)+(32+33)+...+(359+360)
=4+32(1+3)+...+359(1+3)
=4+32.4+...+359.4
=4(1+32+...+359) chia hết cho 4
S=1+3+32+...+360
=(1+3+32)+....+(358+359+360)
=13+...+358(1+3+32)
=13+...+358.13
=13(1+...+358)
a) S=\(1-3+3^2-3^3+...+3^{98}-3^{99}.\)
=\((1-3+3^2-3^3)+...+3^{96}-3^{97}+3^{98}-3^{99}.\)
=\(\left(1-3+3^2-3^3\right)+..+3^{96}\left(1-3+3^2-3^3\right)\)
=(\(1-3+3^2-3^3\))(1+\(3^4+...+3^{92}+3^{96})\)
=-20(1+\(3^4+...+3^{92}+3^{96})\)là bội của -20