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Question Number 9517 by Joel575 last updated on 12/Dec/16

f(1) + f(2) + f(3) + ... + f(n) = n^2 . f(n) n > 1; f(1) = 2016 So, f(2016) = ?

$${f}\left(\mathrm{1}\right)\:+\:{f}\left(\mathrm{2}\right)\:+\:{f}\left(\mathrm{3}\right)\:+\:...\:+\:{f}\left({n}\right)\:=\:{n}^{\mathrm{2}} .\:{f}\left({n}\right) \\ $$ $${n}\:>\:\mathrm{1};\:{f}\left(\mathrm{1}\right)\:=\:\mathrm{2016} \\ $$ $$\mathrm{So},\:{f}\left(\mathrm{2016}\right)\:=\:? \\ $$

Commented bysou1618 last updated on 12/Dec/16

(i) :f(1)+f(2)+...+f(n−1)+f(n)=n^2 f(n) (ii):f(1)+f(2)+...+f(n−1) =(n−1)^2 f(n−1) (i)−(ii)⇒ f(n)=n^2 f(n)−(n−1)^2 f(n−1) ⇔(n^2 −1)f(n)=(n−1)^2 f(n−1) ,(n≠1) ⇔(n+1)f(n)=(n−1)f(n−1) ,(n≠0) ⇔n(n+1)f(n)=(n−1)nf(n−1) g(n)=n(n+1)f(n) ⇒g(n)=g(n−1) ⇒g(n)=g(n−1)=g(n−2)....=g(1)=1×2×2016 ⇒f(n)=((2×2016)/(n(n+1))) f(2016)=((2×2016)/(2016×2017))=(2/(2017))

$$ \\ $$ $$\left({i}\right)\:\::{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+...+{f}\left({n}−\mathrm{1}\right)+{f}\left({n}\right)={n}^{\mathrm{2}} {f}\left({n}\right) \\ $$ $$\left({ii}\right):{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+...+{f}\left({n}−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({n}−\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}−\mathrm{1}\right) \\ $$ $$\left({i}\right)−\left({ii}\right)\Rightarrow \\ $$ $${f}\left({n}\right)={n}^{\mathrm{2}} {f}\left({n}\right)−\left({n}−\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}−\mathrm{1}\right) \\ $$ $$\Leftrightarrow\left({n}^{\mathrm{2}} −\mathrm{1}\right){f}\left({n}\right)=\left({n}−\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}−\mathrm{1}\right)\:\:\:\:\:\:\:,\left({n}\neq\mathrm{1}\right) \\ $$ $$\Leftrightarrow\left({n}+\mathrm{1}\right){f}\left({n}\right)=\left({n}−\mathrm{1}\right){f}\left({n}−\mathrm{1}\right)\:\:\:\:\:\:\:\:,\left({n}\neq\mathrm{0}\right) \\ $$ $$\Leftrightarrow{n}\left({n}+\mathrm{1}\right){f}\left({n}\right)=\left({n}−\mathrm{1}\right){nf}\left({n}−\mathrm{1}\right) \\ $$ $${g}\left({n}\right)={n}\left({n}+\mathrm{1}\right){f}\left({n}\right) \\ $$ $$\Rightarrow{g}\left({n}\right)={g}\left({n}−\mathrm{1}\right) \\ $$ $$\Rightarrow{g}\left({n}\right)={g}\left({n}−\mathrm{1}\right)={g}\left({n}−\mathrm{2}\right)....={g}\left(\mathrm{1}\right)=\mathrm{1}×\mathrm{2}×\mathrm{2016} \\ $$ $$\Rightarrow{f}\left({n}\right)=\frac{\mathrm{2}×\mathrm{2016}}{{n}\left({n}+\mathrm{1}\right)} \\ $$ $$ \\ $$ $${f}\left(\mathrm{2016}\right)=\frac{\mathrm{2}×\mathrm{2016}}{\mathrm{2016}×\mathrm{2017}}=\frac{\mathrm{2}}{\mathrm{2017}} \\ $$

Commented byJoel575 last updated on 12/Dec/16

in (ii) why u wrote (n−1)^2 f(n+1)? is it a typo?

$$\mathrm{in}\:\left({ii}\right)\:\mathrm{why}\:\mathrm{u}\:\mathrm{wrote}\:\left({n}−\mathrm{1}\right)^{\mathrm{2}} \:{f}\left({n}+\mathrm{1}\right)?\:\mathrm{is}\:\mathrm{it}\:\mathrm{a}\:\mathrm{typo}? \\ $$

Commented bysou1618 last updated on 12/Dec/16

oh.... Thank you very much! I fixed it.

$${oh}....\:{Thank}\:{you}\:{very}\:{much}! \\ $$ $${I}\:{fixed}\:{it}. \\ $$

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