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

f:N→R f(1)=2005. and f(1)+f(2)+......+f(n)= n^2 f(n),n>1. Then f(2004)=?

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

Commented byJoel578 last updated on 09/Apr/18

The denomitator form a series 1 3_(⌣) 6_(⌣) 10_(⌣) 15_(⌣) 21_(⌣) , ... +2 +3 +4 +5 +6 Suppose the general 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 After few calculations, you will found a = (1/2), b = (1/2), c = 0 → T_n = (n^2 /2) + (n/2) = (n/2)(n + 1)

$$\mathrm{The}\:\mathrm{denomitator}\:\mathrm{form}\:\mathrm{a}\:\mathrm{series} \\ $$ $$\underset{\smile} {\mathrm{1}\:\:\:\:\:\mathrm{3}}\underset{\smile} {\:\:\:\:\:\:\mathrm{6}}\underset{\smile} {\:\:\:\:\:\:\mathrm{10}}\underset{\smile} {\:\:\:\:\:\:\mathrm{15}}\underset{\smile} {\:\:\:\:\:\:\mathrm{21}},\:... \\ $$ $$\:+\mathrm{2}\:\:+\mathrm{3}\:\:+\mathrm{4}\:\:\:\:\:+\mathrm{5}\:\:\:\:\:+\mathrm{6} \\ $$ $$ \\ $$ $$\mathrm{Suppose}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:{T}_{{n}} \:=\:{an}^{\mathrm{2}} \:+\:{bn}\:+\:{c} \\ $$ $${T}_{\mathrm{1}} \:=\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{1} \\ $$ $${T}_{\mathrm{2}} \:=\:\mathrm{4}{a}\:+\:\mathrm{2}{b}\:+\:{c}\:=\:\mathrm{3} \\ $$ $${T}_{\mathrm{3}} \:=\:\mathrm{9}{a}\:+\:\mathrm{3}{b}\:+\:{c}\:=\:\mathrm{6} \\ $$ $$\mathrm{After}\:\mathrm{few}\:\mathrm{calculations},\:\mathrm{you}\:\mathrm{will}\:\mathrm{found} \\ $$ $${a}\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:{c}\:=\:\mathrm{0} \\ $$ $$\rightarrow\:{T}_{{n}} \:=\:\frac{{n}^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{{n}}{\mathrm{2}}\:=\:\frac{{n}}{\mathrm{2}}\left({n}\:+\:\mathrm{1}\right) \\ $$

Commented byJoel578 last updated on 09/Apr/18

Sir, I think the equation is f(1) + f(2) + ... + f(n) = n^2 f(n)

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

Commented byrahul 19 last updated on 09/Apr/18

Joel sir, suppose we are not able to see the pattern at the first sight. Then how can we find the general term ?

$${Joel}\:{sir},\:{suppose}\:{we}\:{are}\:{not}\:{able}\:{to} \\ $$ $${see}\:{the}\:{pattern}\:{at}\:{the}\:{first}\:{sight}. \\ $$ $${Then}\:{how}\:{can}\:{we}\:{find}\:{the}\:{general} \\ $$ $${term}\:? \\ $$

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

Sorry sir I misunderstood then.

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

Commented byrahul 19 last updated on 09/Apr/18

thank u so much sir!

$${thank}\:{u}\:{so}\:{much}\:{sir}! \\ $$

Answered by Joel578 last updated on 09/Apr/18

f(1) = ((2005)/1) f(2) = ((2005)/3) f(3) = ((2005)/6) f(4) = ((2005)/(10)) f(5) = ((2005)/(15)) ⋮ f(n) = ((2005)/((n/2)(n + 1))) f(2004) = ((2005)/(1002 . 2005)) = (1/(1002))

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

Commented byrahul 19 last updated on 09/Apr/18

thank u sir!

$${thank}\:{u}\:{sir}! \\ $$

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