答案113.
設(shè)第n次有a_n個(gè)球. 則有遞推關(guān)系
a_n=a_{n-1}-n+3n=a_{n-1}+2n.
故a_n
=a_{n-1}+2n
=a_{n-2}+2(n-1)+2n
=a_{n-3}+2(n-2)+2(n-1)+2n
=...
=a_0+2*(1+2+3+...+n)
=3+n(n+1)
=n^2+n+3.
故a_{10}=113.