Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 209718 by Ismoiljon_008 last updated on 19/Jul/24

Answered by mr W last updated on 19/Jul/24

a_n =(1/(n!×(n+2)))=(1/((n+1)!))−(1/((n+2)!)) Σ_(n=1) ^(2023) a_n =((1/(2!))−(1/(3!)))+((1/(3!))−(1/(4!)))+...+((1/(2024!))−(1/(2025!)) =(1/2)−(1/(2025!)) ⇒C)

an=1n!×(n+2)=1(n+1)!−1(n+2)!∑2023n=1an=(12!−13!)+(13!−14!)+...+(12024!−12025!=12−12025!⇒C)

Commented by Ismoiljon_008 last updated on 19/Jul/24

thank you Mr W I appreciate you

thankyouMrWIappreciateyou

Terms of Service

Privacy Policy

Contact: info@tinkutara.com