주기 함수 합

조건 (가), (나)에서 $f ( 3 ) = f ( 1 + 2 ) = 2 f ( 1 ) = 10$에서 $f ( 1 ) = 5$ ∴ $\displaystyle \sum\limits _{n=1} ^{5} f ( 2 n - 1 ) = 5 ( 1 + 2 + 2 ^{2} + 2 ^{3} + 2 ^{4} )$$= 5 \times 31$ $f ( 2 ) = a$라 하면 조건(나)에서 $\displaystyle \sum\limits _{n=1} ^{5} f ( 2 n ) = a ( 1 + 2 + 2 ^{2} + 2 ^{3} + 2 ^{4} )$$= a \times 31$ 조건 (다)에 의해서 $\displaystyle \sum\limits _{n=1} ^{100} f ( n )$$\displaystyle = 10 \times \sum\limits _{n=1} ^{10} f ( n )$ $\displaystyle = 10 \times \sum\limits _{n=1} ^{5} \left\{ f ( 2 n - 1 ) + f ( 2 n ) \right\}$ $= 10 \times ( a + 5 ) \times 31$ $= 2170$ ∴ $a = 2$ $f ( 100 )$$= f ( 10 )$$= 2 ^{4} f ( 2 )$$= 2 ^{4} \times a$$= 32$