3²(1+3+...+3^2008)=9*(3^2009-1)/(3-1)=9*(3^2009-1)/2
cu substitutia 3^7=u
avem
(1+3⁷+3¹⁴+...+3²⁰⁰² ) =1+u+u²+...+u^286= (u^287-1)/(u-1)=(3^2009 -1)/(3^7-1)
egalitatea din data problemeidevine
9(3^2009-1)/2=a * b² *(3^2009-1)/(3^7-1)*
simplificand cu 3^2009 -1, rezulta
9/2=a*b²/(3^7-1)
9/2=a*b²/2186
9=a*b²/1093
9*1093=a*b²
1093 *3²=ab²
1093 numar prim
a=1093 ..b=3
a=1093
a+b^6+2b^4+b³-b^0=1093+729+2*81+27-1=2010 adevarat
