Cho p = 1+3^1+3^2 + ............................+3^99
a, Chứng minh rằng p:4
b, chứng mình rằng p:40
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C = (1 + 3 + 3^2 + 3^3) + ......... + (3^8 + 3^9 + 3^10 + 3^11)
C = 1.40 + .............. + 3^8.40
= 40.(1 + 3^4 + ...... + 3^8)
Chia hết cho 40
A=(3^0+3^1+3^2+3^3)+(3^4+3^5+3^6+3^7)+...+(3^2009+3^2010+3^2011+3^2012)
A=40+3^4*(1+3+3^2+3^3)+...+3^2009*(1+3+3^2+3^3)
A-1=40+80*40+...+3^2009*40
A-1=40*(1+80+..+3^2009)
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
A= 1 +(3^1+3^2+3^3+3^4)+..............................+(3^2009+3^2010+3^2011+3^2012)
A=1+120+................................+3^2009*(3^1+3^2+3^3+3^4)
A=1+(1+.....................+3^2009)*120
Vì 120 chia hết cho 40
suy ra (1+..........................+3^2009) chia hết cho 40
suy ra A chia 40 dư 1
suy ra A-1 chia hết cho 40
p = (1 + 31) + (32 + 33) + ...+ (398 + 399) = 4.1 + 32.(1+ 31) + ...+ 398.(1+ 31) = 4.1 + 32.4 + ....+ 398.4
= 4. (1 + 32 + 34 + ...+ 398) chia hết cho 4
=> p chia hết cho 4
p = (1+ 31+ 32 + 33) + (34 + 35 + ...+ 37) + ...+ (396 + 397 + 398 + 399)
= 40 + 34.(1 + 31 + 32+ 33) + ....+ 396. (1+31 + 32 + 33)
= 40 + 34. 40 + ....+ 396.40 = 40.(1 + 34 + ...+ 396) chia hết cho 40
=> p chia hết cho 40