"Sum" of the functions are interesting

Algebra Level 3
  • f k ( x ) f_k (x) has three roots and they are k k , k -k and 0 0 , where k k is a real positive integer 1 \geq 1 . Which of the followings (in terms of n n ) is the value of x x when f 1 ( x ) + f 2 ( x ) + + f n ( x ) = 0 f_1 (x)+f_2 (x)+\cdots+f_n (x)=0 ?
  • Assume all f k ( x ) f_k (x) has equal real constant d d , such that f k ( x ) = d ( x ) ( x k ) ( x + k ) f_k (x)=d(x)(x-k)(x+k) where d d is not 0 0
  • Given that x > 0 x>0 and n = 24 n=24 ,
  • express your answer in the form a b c \dfrac{a \sqrt b}c , where a , b a,b and c c are positive integers with a , c a,c coprime and b b square-free.

Submit your answer in the form a + b + c a+b+c .


The answer is 47.

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1 solution

James Wong
Mar 25, 2016
  • f 1 ( x ) = d x ( x 2 1 ) f_1 (x)=dx(x^2-1)
  • f 2 ( x ) = d x ( x 2 4 ) f_2 (x)=dx(x^2-4) and so on.
  • Sum of f k ( x ) f_k (x) from f 1 ( x ) f_1 (x) up to f n ( x ) f_n (x) = d x ( n x 2 1 2 2 2 3 2 4 2 . . . n 2 ) =dx(nx^2-1^2-2^2-3^2-4^2-...-n^2)
  • Which is equal to d x ( n x 2 ( 1 2 + 2 2 + 3 2 + . . . + n 2 ) ) dx(nx^2-(1^2+2^2+3^2+...+n^2))
  • Ultimately equal to d x ( n x 2 n ( n + 1 ) ( 2 n + 1 ) / 6 ) dx(nx^2-n(n+1)(2n+1)/6)
  • So, as x x cannot be zero, x = ( ( n + 1 ) ( 2 n + 1 ) / 6 ) x=√((n+1)(2n+1)/6)
  • Substitute n = 24 , x = ( 35 6 ) / 6 n=24, x=(35√6)/6
  • 35 + 6 + 6 = 47 35+6+6=47

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