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Question Number 33032 by rahul 19 last updated on 09/Apr/18

$$f:N\to R$$ $$f\left(1\right)=2005.$$ $$and$$ $$f\left(1\right)+f\left(2\right)+......+f\left(n\right)=n^{2}f\left(n\right),n>1.$$ $$Thenf\left(2004\right)=?$$

Commented byJoel578 last updated on 09/Apr/18

$$\mathrm{The}\mathrm{denomitator}\mathrm{form}\mathrm{a}\mathrm{series}$$ $$\underset{⌣}{13}\underset{⌣}{6}\underset{⌣}{10}\underset{⌣}{15}\underset{⌣}{21},...$$ $$+2+3+4+5+6$$ $$$$ $$\mathrm{Suppose}\mathrm{the}\mathrm{general}\mathrm{term}{T}_n={an}^2+bn+c$$ $$T_1=a+b+c=1$$ $$T_2=4a+2b+c=3$$ $$T_3=9a+3b+c=6$$ $$\mathrm{After}\mathrm{few}\mathrm{calculations},\mathrm{you}\mathrm{will}\mathrm{found}$$ $$a=\frac{1}{2},b=\frac{1}{2},c=0$$ $$\to T_n=\frac{n^2}{2}+\frac{n}{2}=\frac{n}{2}(n+1)$$

Commented byJoel578 last updated on 09/Apr/18

$$\mathrm{Sir},\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{equation}\mathrm{is}$$ $$f\left(1\right)+f\left(2\right)+...+f\left(n\right)=n^{2}f\left(n\right)$$

Commented byrahul 19 last updated on 09/Apr/18

$$Joelsir,supposewearenotableto$$ $$seethepatternatthefirstsight.$$ $$Thenhowcanwefindthegeneral$$ $$term?$$

Commented byRasheed.Sindhi last updated on 09/Apr/18

$$\mathrm{Sorry}\mathrm{sir}\mathrm{I}\mathrm{misunderstood}\mathrm{then}.$$

Commented byrahul 19 last updated on 09/Apr/18

$$thankusomuchsir!$$

Answered by Joel578 last updated on 09/Apr/18

$$f\left(1\right)=\frac{2005}{1}$$ $$f\left(2\right)=\frac{2005}{3}$$ $$f\left(3\right)=\frac{2005}{6}$$ $$f\left(4\right)=\frac{2005}{10}$$ $$f\left(5\right)=\frac{2005}{15}$$ $$⋮$$ $$f\left(n\right)=\frac{2005}{\frac{n}{2}(n+1)}$$ $$f\left(2004\right)=\frac{2005}{1002.2005}=\frac{1}{1002}$$

Commented byrahul 19 last updated on 09/Apr/18

$$thankusir!$$

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