It is proved that (1) when n= 1, the proposition is clearly established.

When n=2, f(2)=(2×2+7)? 3? +9= 12×9=36×3 proposition holds.

(2) Suppose that the proposition holds when n=k [k∈N* k≥2].

That is f(k)=(2k+7)? 3 k+9 is divisible by 36, so when n=k+ 1

f(k+ 1)=[2(k+ 1)+7]？ 3^(k+ 1)+9=[(2k+7)？ 3^(k+ 1)+27]+2×3^(k+ 1)- 18

=3? [(2k+7)？ 3^k+9]+6×3^k- 18=3[(2k+7)？ 3^k+9]+ 18？ [3^(k- 1)- 1]

And 3 (k-1)-1= (1+2) (k-1)-1.

The first term in the expansion of (1+2) (k- 1) is 1, and the other terms each contain at least 1 a power term of 2, so

Can 3 (k- 1)- 1 be divisible by 2 18? [3 (k- 1)- 1] is divisible by 36.

3[(2k+7)？ 3^k+9]+ 18？ [3 (k- 1)- 1] is divisible by 36, that is, when n=k+ 1, the proposition holds.