设数列{bn}满足b1=2009,对n＞1,b1+b2+.+bn=(1+2+.+n)bn,则b2008=

1+b2+.+bn=(1+2+.n)bn=n(n+1)bn/2b1+b2+b3+...+bn-1=bn(n+2)(n-1)/2 b1+..+bn-2=bn-1(n+1)(n-2)/2 bn-1=bn(n+2)(n-1)/2-bn-1(n+1)(n-2)/2 bn-1(n-1)n/2=bn(n+2)(n-1)/2 bn/bn-1=n/(n+2) bn=bn/bn-1*bn-1/bn-2*.*b1 =n/(n+2)*(n-1)/(n+1)*.*2/4*2009 =6*2009/(n+1)(n+2) =2009*6/(n+1)(n+2) b2008=6/2010=3/1005

令Sn=b1+b2+......+bn 因为1+2+......+n=n(n+1)/2 所以由题意可得Sn=[n(n+1)/2]*bn 所以S 两式相减可得Sn-S=[n(n+1)*bn-n(n-1)*b]/2 所以2bn=n(n+1)*bn-n(n-1)*b 2bn=(n+n)*bn-(n-n)*b (n-n)*...