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Question Number 90259 by john santu last updated on 22/Apr/20
a function f is defined on the positive integers satisfies f(1) = 1002 , & f(1)+f(2)+f(3)+ ... +f(n) = n^2 f(n) . find f(2003)
$${a}\:{function}\:{f}\:{is}\:{defined}\:{on}\: \\ $$$${the}\:{positive}\:{integers}\:{satisfies}\: \\ $$$${f}\left(\mathrm{1}\right)\:=\:\mathrm{1002}\:,\:\&\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+ \\ $$$$…\:+{f}\left({n}\right)\:=\:{n}^{\mathrm{2}} \:{f}\left({n}\right)\:. \\ $$$${find}\:{f}\left(\mathrm{2003}\right)\: \\ $$
Answered by mr W last updated on 22/Apr/20
Σ_(k=1) ^n f(k)=n^2 f(n) ...(i) Σ_(k=1) ^(n+1) f(k)=(n+1)^2 f(n+1) Σ_(k=1) ^n f(k)+f(n+1)=(n+1)^2 f(n+1) ...(ii) (ii)−(i): f(n+1)=(n+1)^2 f(n+1)−n^2 f(n) ⇒f(n+1)=(n/(n+2))f(n) ⇒f(n)=((n−1)/(n+1))f(n−1) =(((n−1)(n−2)...2×1)/((n+1)n(n−1)...4×3))f(1) ⇒f(n)=(2/((n+1)n))f(1) f(2003)=(2/(2004×2003))×1002=(1/(2003))
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({k}\right)={n}^{\mathrm{2}} {f}\left({n}\right)\:\:\:…\left({i}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}{f}\left({k}\right)=\left({n}+\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}+\mathrm{1}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left({k}\right)+{f}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}+\mathrm{1}\right)\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${f}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}+\mathrm{1}\right)−{n}^{\mathrm{2}} {f}\left({n}\right) \\ $$$$\Rightarrow{f}\left({n}+\mathrm{1}\right)=\frac{{n}}{{n}+\mathrm{2}}{f}\left({n}\right) \\ $$$$\Rightarrow{f}\left({n}\right)=\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}}{f}\left({n}−\mathrm{1}\right) \\ $$$$=\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\mathrm{2}×\mathrm{1}}{\left({n}+\mathrm{1}\right){n}\left({n}−\mathrm{1}\right)…\mathrm{4}×\mathrm{3}}{f}\left(\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left({n}\right)=\frac{\mathrm{2}}{\left({n}+\mathrm{1}\right){n}}{f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{2003}\right)=\frac{\mathrm{2}}{\mathrm{2004}×\mathrm{2003}}×\mathrm{1002}=\frac{\mathrm{1}}{\mathrm{2003}} \\ $$
Commented by john santu last updated on 22/Apr/20
the relation you obtained between f(n) and f(n−1) is correct. by the way, in the step before with f(n+1) you wrote f(n−1) instead of f(n). the relation you obtained f(n) and f(1) is incorrect . it should be a 2 and not 6. the quotient of product doesn′t seem right. so i′m assuming the error came from there.
$${the}\:{relation}\:{you}\:{obtained}\:{between}\: \\ $$$${f}\left({n}\right)\:{and}\:{f}\left({n}−\mathrm{1}\right)\:{is}\:{correct}. \\ $$$${by}\:{the}\:{way},\:{in}\:{the}\:{step}\:{before} \\ $$$${with}\:{f}\left({n}+\mathrm{1}\right)\:{you}\:{wrote}\:{f}\left({n}−\mathrm{1}\right) \\ $$$${instead}\:{of}\:{f}\left({n}\right).\:{the}\:{relation}\: \\ $$$${you}\:{obtained}\:{f}\left({n}\right)\:{and}\:{f}\left(\mathrm{1}\right)\:{is} \\ $$$${incorrect}\:.\:{it}\:{should}\:{be}\:{a}\:\mathrm{2}\:{and} \\ $$$${not}\:\mathrm{6}.\:{the}\:{quotient}\:{of}\:{product}\: \\ $$$${doesn}'{t}\:{seem}\:{right}.\:{so}\:{i}'{m}\: \\ $$$${assuming}\:{the}\:{error}\:{came}\: \\ $$$${from}\:{there}. \\ $$$$ \\ $$
Commented by john santu last updated on 22/Apr/20
the answer is (1/(2003))
$${the}\:{answer}\:{is}\:\frac{\mathrm{1}}{\mathrm{2003}} \\ $$
Commented by mr W last updated on 22/Apr/20
now fixed. thanks!
$${now}\:{fixed}.\:{thanks}! \\ $$

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