令p=q=n，得f2（n）=f（2n）．
原式=
| 2f2(1) |
| f(1) |
| 2f(4) |
| f(3) |
| 2f(6) |
| f(5) |
| 2f(8) |
| f(7) |
=2f（1）+
| 2f(1)f(3) |
| f(3) |
| 2f(1)f(5) |
| f(5) |
| 2f(1)f(7) |
| f(7) |
=8f（1）=24．
故答案为：24．
| f2(1)+f(2) |
| f(1) |
| f2(2)+f(4) |
| f(3) |
| f2(3)+f(6) |
| f(5) |
| f2(4)+f(8) |
| f(7) |
| 2f2(1) |
| f(1) |
| 2f(4) |
| f(3) |
| 2f(6) |
| f(5) |
| 2f(8) |
| f(7) |
| 2f(1)f(3) |
| f(3) |
| 2f(1)f(5) |
| f(5) |
| 2f(1)f(7) |
| f(7) |