令a=33···3(n个3)=99···9(n个9)/3=(10ⁿ-1)/3,
则原式=a·(a+1)
=[(10ⁿ-1)/3]·[(10ⁿ-1)/3+1]=[(10ⁿ-1)/3]·[(10ⁿ+2)/3]
=(10²ⁿ+10ⁿ-2)/9
=(10²ⁿ-1)/9+(10ⁿ-1)/9
=11···1(2n个1)+11···1(n个1)
=11···122···2(前边是n个1,后边是n个2)