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wu :: forums    riddles    medium (Moderators: Icarus, Eigenray, william wu, towr, Grimbal, SMQ, ThudnBlunder)    f(2007) « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: f(2007)  (Read 419 times) perash Newbie     Posts: 4 f(2007)   « on: Jan 4th, 2007, 11:25am » Quote Modify if  f(1)=2007 n is integers f(1)+f(2)+.......+f(n)=(n^2).f(n) and n>= 1 find f(2007) IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: f(2007)   « Reply #1 on: Jan 4th, 2007, 12:32pm » Quote Modify f(n)  = 2007 / (1/2 n(n+1))     f(2007) = 2/2008 = 1/1004 IP Logged Wikipedia, Google, Mathworld, Integer sequence DB balakrishnan Junior Member     Gender: Posts: 92 Re: f(2007)   « Reply #2 on: Jan 6th, 2007, 10:31am » Quote Modify f(1)+f(2)+..f(n)=n^2 f(n) f(1)+..f(n-1)=(n-1)^2 f(n-1) subtracting the 2,we get f(n)=n^2 f(n)-(n-1)^2 f(n-1) which gives (n^2-1) f(n) =(n-1)^2 f(n-1) or (n+1) f(n)=(n-1) f(n-1) or f(n)=(n-1)/(n+1) f(n-1)   which means f(n)=(n-1)/(n+1)*(n-2)/(n)*(n-3)/(n-1)*...(1/3)*f(1) or f(n)=[(n-1)!]/[(n+1)!] *2 * f(1)=2*f(1)/[n(n+1)] which gives f(2007)=2/2008=1/1004 IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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