f-1-2025-1-n-f-k-n-2-f-n-f-2025-

Question Number 223400 by behi834171 last updated on 23/Jul/25

$$\boldsymbol{{f}}\left(\mathrm{1}\right)=\mathrm{2025} \\ $$$$\underset{\mathrm{1}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}{f}}\left(\boldsymbol{{k}}\right)=\boldsymbol{{n}}^{\mathrm{2}} .\boldsymbol{{f}}\left(\boldsymbol{{n}}\right) \\ $$$$\boldsymbol{{f}}\left(\mathrm{2025}\right)=? \\ $$

Answered by mr W last updated on 24/Jul/25

$${S}_{{n}} ={n}^{\mathrm{2}} {f}\left({n}\right) \\ $$$${S}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}+\mathrm{1}\right)={S}_{{n}} +{f}\left({n}+\mathrm{1}\right) \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{2}} {f}\left({n}+\mathrm{1}\right)={n}^{\mathrm{2}} {f}\left({n}\right)+{f}\left({n}+\mathrm{1}\right) \\ $$$$\left({n}+\mathrm{2}\right){f}\left({n}+\mathrm{1}\right)={nf}\left({n}\right) \\ $$$$\frac{{f}\left({n}+\mathrm{1}\right)}{{f}\left({n}\right)}=\frac{{n}}{{n}+\mathrm{2}} \\ $$$$\frac{{f}\left({n}\right)}{{f}\left({n}−\mathrm{1}\right)}=\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$… \\ $$$$\frac{{f}\left(\mathrm{2}\right)}{{f}\left(\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{{f}\left({n}\right)}{{f}\left(\mathrm{1}\right)}=\frac{\mathrm{2}×\mathrm{1}}{\left({n}+\mathrm{1}\right){n}} \\ $$$$\Rightarrow{f}\left({n}\right)=\frac{\mathrm{2}{f}\left(\mathrm{1}\right)}{\left({n}+\mathrm{1}\right){n}} \\ $$$${f}\left(\mathrm{2025}\right)=\frac{\mathrm{2}×\mathrm{2025}}{\mathrm{2026}×\mathrm{2025}}=\frac{\mathrm{1}}{\mathrm{1013}}\:\checkmark \\ $$

Commented by behi834171 last updated on 24/Jul/25

$${thank}\:{you}\:{very}\:{much}\:{dear}\:{master}. \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com