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S=2+22+23+24+....+299+2100
=(2+22+23) + ( 24+25+26) + ......+ ( 288+299+2100)
= 2.14+24.14+....+288.14
= 14.( 2+24+....+288) Chia hết cho 14
Vậy S chia hết cho 14
\(S=5+5^2+5^3+5^4+...+5^{2012}\)
\(S=\left(5+5^3\right)+\left(5^2+5^4\right)+...+\left(5^{2010}+5^{2012}\right)\)
\(S=\left(5+5^3\right)+5\left(5+5^3\right)+...+5^{2009}\left(5+5^3\right)\)
\(S=130+5\cdot130+...+5^{2009}\cdot130\)
\(S=65\cdot2+5\cdot65\cdot2+...+5^{2009}\cdot65\cdot2\)
\(S=65\left(2+5\cdot2+...+5^{2009}\cdot2\right)⋮65\) (đpcm)
=))
a) Ta có:
\(S=1+2+2^2+...+2^{119}\)
\(S=\left(1+2+2^2+2^3\right)+\left(2^3+2^4+2^5+2^6\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}\right)\)
\(S=\left(1+2+2^2+2^3\right)+2^3\cdot\left(1+2+2^2+2^3\right)+...+2^{116}\cdot\left(1+2+2^2+2^3\right)\)
\(S=15+15\cdot2^3+...+15\cdot2^{116}\)
\(S=15\cdot\left(1+2^3+...+2^{116}\right)\) chia hết cho 5
b) \(S=1+2+2^2+...+2^{119}\)
\(\Rightarrow2S=2+2^2+2^3+...+2^{120}\)
\(\Rightarrow2S-S=\left(2+2^2+...+2^{120}\right)-\left(1+2+...+2^{119}\right)\)
\(\Leftrightarrow S=2^{120}-1\)
\(\Leftrightarrow2^n=S+1=2^{120}\)
\(\Rightarrow n=120\)
Ta có: S=1+4+42+43+…+459
=>S=(1+4+42+43)+…+(456+457+458+459)
=>S=(1+4+42+43)+…+456.(1+4+42+43)
=>S=85+…+456.85
=>S=(1+…+456).85 chia hết cho 85
=>S chia hết cho 85
Bài 1:
a) \(\left|2y+1\right|=7\)
\(\Rightarrow2y+1=7\) hoặc \(2y+1=-7\)
\(\Rightarrow2y=6\) hoặc \(2y=-8\)
\(\Rightarrow y=3\) hoặc \(y=-4\)
Vậy................
b) \(\left|y-8\right|-15=22\)
\(\left|y-8\right|=37\)
\(\Rightarrow y-8=37\) hoặc \(y-8=-37\)
\(\Rightarrow y=45\) hoặc \(y=-29\)
Vậy \(y\in\left\{45;-29\right\}\)
Ta có : S=4+32+33+...+3223
=1+3+32+33+...+3223
=(1+32+34+36)+(3+33+35+37)+...+(3217+3219+3221+3223)
=1(1+32+34+36)+3(1+32+34+36)+...+3217(1+32+34+36)
=1.820+3.820+...+3217.820
Vì 820\(⋮\)41 nên 1.820+3.820+...+3217.820\(⋮\)41
hay S\(⋮\)41
Vậy S\(⋮\)41.