數(shù)列an各項(xiàng)為整數(shù),且(n-1)(a(n+1)+2)=(n+1)an,a2005是7的倍數(shù),an=An^2+Bn+C,則最小的正整數(shù)A=?
數(shù)列an各項(xiàng)為整數(shù),且(n-1)(a(n+1)+2)=(n+1)an,a2005是7的倍數(shù),an=An^2+Bn+C,則最小的正整數(shù)A=?
數(shù)學(xué)人氣:396 ℃時(shí)間:2020-04-05 17:25:55
優(yōu)質(zhì)解答
由于(n-1)[a(n+1)+2]≡0(mod n+1)-2(C+2)≡0 (mod n+1)C=-2an=An^2+Bn+C , 帶入條件A(n+1)+B(n+1)-2(n+1)=0A+B=2又an都是整數(shù)所以 A=x/2 ,B=y/2 x+y=4 ,x,y為整數(shù)2005≡3 (mod 7)所以(2x+3y-4)/2 ≡0(mod 7)8-x ...
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