Function in Bulk

Algebra Level 4

Let f ( x ) = a x + b f(x)=ax+b , where a a and b b are integers, f 2 ( 0 ) = f ( f ( 0 ) ) = 0 f^2 (0) =f(f(0))=0 and f 3 ( 2 ) = f ( f ( f ( 2 ) ) ) = 9 f^3 (2) = f(f(f(2)))=9 .

Find f 4 ( 1 ) + f 4 ( 2 ) + f 4 ( 3 ) + ⋯ + f 4 ( 2016 ) f^4(1)+ f^4(2)+ f^4(3)+\cdots+ f^4(2016)


The answer is 2033136.

Choi Chakfung by choi chakfung , 28, Hong Kong

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1 solution

f ( 0 ) = b f(0)=b . f ( f ( 0 ) ) = f ( b ) = 0 f(f(0))=f(b)=0 . This simplifies to a b + b = 0 ab+b=0 or a = − 1 a=-1 .

f ( 2 ) = b − 2 f(2)=b-2 . f ( b − 2 ) = 2 f(b-2)=2 . Therefore f ( f ( f ( 2 ) ) ) = f ( 2 ) = b − 2 = 9 f(f(f(2)))=f(2)=b-2=9 . Therefore b = 11 b=11 .

Therefore f ( x ) = 11 − x f(x)=11-x .

As f ( f ( f ( f ( x ) ) ) ) = x f(f(f(f(x))))=x for all x x , the value of the expression is 1 + 2 + … + 2016 = 2016 × 2017 2 = 2033136 1+2+\ldots+2016=\dfrac{2016 \times 2017} 2 = \boxed{2033136}

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There is another option, b = 0 b=0 . You didn't show why this is not possible

Hung Woei Neoh - 5 years ago

Please explain

Aditya Sky - 4 years ago

b ≠ 0 b≠0 as it would lead to a = ( 4.5 ) 1 3 a= (4.5)^{\frac{1}{3}} which is not an integer.

Aditya Sky - 4 years ago

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