Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 97136 by I want to learn more last updated on 06/Jun/20

Evaluate (3/(1! + 2! + 3!)) + (4/(2! + 3! + 4!)) + ... + ((2001)/(1999! + 2000! + 2001!))

$$\mathrm{Evaluate} \\ $$$$\:\:\frac{\mathrm{3}}{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!}\:\:+\:\:\frac{\mathrm{4}}{\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!}\:\:+\:\:...\:+\:\:\frac{\mathrm{2001}}{\mathrm{1999}!\:\:+\:\:\mathrm{2000}!\:\:+\:\:\mathrm{2001}!} \\ $$

Commented by 06122004 last updated on 06/Jun/20

∗Yechim∗ n!+(n+1)!+(n+2)!=n!(1+(n+1)+(n+1)(n+2))= =n!∗(n+2)^2 ⊛ (3/(1!∗3^2 ))+(4/(2!∗4^2 ))+...+((2001)/(1999!∗2001^2 ))= =(1/(1!∗3))+(1/(2!∗4))+...+(1/(1999!∗2001))= =(2/(3!))+(3/(4!))+...+((2000)/(2001!))= =((3−1)/(3!))+((4−1)/(4!))+...+((2001−1)/(2001!))= =(1/(2!))−(1/(3!))+(1/(3!))−(1/(4!))+...+(1/(2000!))−(1/(2001!))= =(1/(2!))−(1/(2001!))=((((2001!)/2)−1)/(2001!))=((2001!−2)/(2∗2001!)) ▲ ∗Javob∗ ((2001!−2)/(2∗2001!)) Abdullaxatov Jasurbek

$$\ast{Yechim}\ast \\ $$$${n}!+\left({n}+\mathrm{1}\right)!+\left({n}+\mathrm{2}\right)!={n}!\left(\mathrm{1}+\left({n}+\mathrm{1}\right)+\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\right)= \\ $$$$={n}!\ast\left({n}+\mathrm{2}\right)^{\mathrm{2}} \:\:\circledast \\ $$$$\frac{\mathrm{3}}{\mathrm{1}!\ast\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{2}!\ast\mathrm{4}^{\mathrm{2}} }+...+\frac{\mathrm{2001}}{\mathrm{1999}!\ast\mathrm{2001}^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}!\ast\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}!\ast\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{1999}!\ast\mathrm{2001}}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+...+\frac{\mathrm{2000}}{\mathrm{2001}!}= \\ $$$$=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{4}−\mathrm{1}}{\mathrm{4}!}+...+\frac{\mathrm{2001}−\mathrm{1}}{\mathrm{2001}!}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}+...+\frac{\mathrm{1}}{\mathrm{2000}!}−\frac{\mathrm{1}}{\mathrm{2001}!}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{2001}!}=\frac{\frac{\mathrm{2001}!}{\mathrm{2}}−\mathrm{1}}{\mathrm{2001}!}=\frac{\mathrm{2001}!−\mathrm{2}}{\mathrm{2}\ast\mathrm{2001}!}\:\:\blacktriangle \\ $$$$\ast{Javob}\ast\:\frac{\mathrm{2001}!−\mathrm{2}}{\mathrm{2}\ast\mathrm{2001}!} \\ $$$${Abdullaxatov}\:{Jasurbek} \\ $$

Commented by I want to learn more last updated on 06/Jun/20

Thanks sir.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by 06122004 last updated on 06/Jun/20

ok!

$${ok}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com