MHB Find f(1996): f(1)=1996, f(1)+f(2)+\cdots+f(n)=n^2f(n)

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The function f is defined such that f(1) equals 1996, and the relationship f(1) + f(2) + ... + f(n) = n^2f(n) holds. By manipulating the equation, it is derived that f(n) can be expressed as f(n) = (n-1)/(n+1) * f(n-1). This recursive relationship leads to the calculation of f(1996) as a product involving decreasing integers and the initial value f(1). Ultimately, f(1996) is determined to be 2/1997.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
If $$f(1)=1996$$, $$f(1)+f(2)+\cdots+f(n)=n^2f(n)$$, find $$f(1996)$$.
 
Mathematics news on Phys.org
anemone said:
If $$f(1)=1996$$, $$f(1)+f(2)+\cdots+f(n)=n^2f(n)$$, find $$f(1996)$$.
let f(1)=k=1996
$$f(1)+f(2)+\cdots+f(n-1)=n^2f(n)-f(n)=(n-1)^2f(n-1)$$
$\therefore f(n)=\dfrac{n-1}{n+1}\times f(n-1)$
$ f(1996)=\dfrac {(k-1)(k-2)(k-3)-------(3)(2)(1)}{{(k+1)}(k)(k-1)----(5)(4)(3)}\times f(1)=\dfrac {2}{k+1}$
$=\dfrac {2}{1997}$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top