Question

已知数列an的前n项和为Sn,f(x)=(2x-1)/(x+1),an=log2f(n+1)/f(n)则S2011=

Answer 1

an=log2[f(n+1)/f(n)]
=log2[f(n+1)]-log2[f(n)]
=log2[(2(n+1)-1)/((n+1)+1)]-log2[(2n-1)/(n+1)]
=log2[2(n+1)-1] -log2[(n+1)+1] -log2(2n-1) +log2(n+1)
=log2(n+1) -log2[(n+1)+1] -{log2(2n-1) -log2[2(n+1)-1]}
S2011=log2(2)-log2(3)+log2(3)-log2(4)+...+log2(2012)-log2(2013)
-[log2(2×1-1)-log2(2×2-1)+log2(2×2-1)-log2(2×3-1)+...+log2(2×2011-1)-log2(2×2012-1)]
=log2(2) -log2(2013)-[log2(1)-log2(4023)]
=log2(2)-log2(2013)-log2(1)+log2(4023)
=1-log2(2013)+0+log2(4023)
=1+ log2(4023) -log2(2013)