答案113.设第n次有a_n个球. 则有递推关系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.
105
3的10次方加20
113
