Oscillation of Invariants
Given that n is a positive integer and k , p are real constants such that f ( n ) = k − p + f ( n − 1 ) ( k + p ) 2 and f ( 1 ) = a .
Find the value of f ( 2 0 1 7 ) f ( 1 ) + f ( 2 0 1 4 ) f ( 4 ) + f ( 2 0 1 1 ) f ( 7 ) + ⋯ + f ( 4 ) f ( 2 0 1 4 ) + f ( 1 ) f ( 2 0 1 7 ) .
The answer is 673.
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1 solution
k = − p is impossible here. A careful observation of the recurrence formula would reveal that, we have, p + f(n-1) as the denominator of a fraction on the right hand side which can not be 0 for f(n) to be real so, f ( n − 1 ) = − p . Assuming k=-p gives all f ( n ) = k = f ( n − 1 ) = − p which is a contradiction. Also that if all f(n)=k, then all f(n)=-p therefore, f(n)+p=0 and you can not divide by 0. So the possibility that k=-p gets eliminated. Brilliant always takes numerical values as answer for non multiple choice questions so it is easy to guess the answer to be independent of a,k and p but this is not the case. However the better option is to think about how the answer can be achieved instead of making assumption about what the answer can possibly be. Its also interesting to get a back again as the fourth term from the third one by using some common algebra. Use of algebra can show that the recurrence formula yields f(n)=f(n+3)=.....for all the other values of k and p.
Not a solution.
As answer is independent of k and p , Let k = − p
This gives f ( 1 ) = a and all other f(n)'s f ( n ) = k .
Then our sum is 6 7 1 + k / a + a / k . To find total number of terms, use properties of n t h term of AP
Lastly let a = k (As answer is independent of a as well). We have our answer as 673
Yes, the 3rd term does pose a problem but it does not affect the answer.