Question-172448 Tinku Tara June 4, 2023 Algebra 0 Comments FacebookTweetPin Question Number 172448 by Shrinava last updated on 27/Jun/22 Answered by Jamshidbek last updated on 28/Jun/22 Let1!!0!+3!!1!+5!!2!+…+(2n+1)!!n!=anandan+1−an=(2n+3)!!(n+1)!Ω=∑∞n=0an⋅2n+2(2n+3)!!=∑∞n=0an⋅(1(2n+1)!!−1(2n+3)!!)==∑∞n=0(an(2n+1)!!−an(2n+3)!!)=∑∞n=0an(2n+1)!!−∑∞n=0an(2n+3)!!(1)∑∞n=0an(2n+1)!!=a0+∑∞n=1an(2n+1)!!=a0+∑∞n=0an+1(2n+3)!!(2)From(1)and(2)Ω=∑∞n=0an(2n+1)!!−∑∞n=0an(2n+3)!!=a0+∑∞n=0an+1(2n+3)!!−∑∞n=0an(2n+3)!!==a0+∑∞n=0an+1−an(2n+3)!!=a0+∑∞n=0(2n+3)!!(n+1)!⋅(2n+3)!!=a0+∑∞n=01(n+1)!==a0+∑∞n=11n!⇒a0=1⇒Ω=1+∑∞n=11n!ex=∑∞n=0xnn!=1+∑∞n=1xnn!Putx=1e=1+∑∞n=11n!⇒Ω=eAnswer:eTelegram:@math_undergraduate= Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-172450Next Next post: Question-172449 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.