find-the-value-of-n-1-1-n-4n-2-1-2- Tinku Tara June 3, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 74351 by mathmax by abdo last updated on 22/Nov/19 findthevalueof∑n=1+∞(−1)n(4n2−1)2 Commented by ~blr237~ last updated on 23/Nov/19 letnameditSS=∑∞n=1(−1)n[(2n+1)(2n−1)]2=∑∞n=1(−1)n2[(2n+1)−(2n−1)(2n+1)(2n−1)]22S=∑∞n=1(−1)n[12n−1−12n+1]22S=∑∞n=1(−1)n[1(2n−1)2−2(2n+1)(2n−1)+1(2n+1)2]2S=∑∞n=1(−1)n(2n−1)2−∑∞n=12(2n+1)(2n−1)+∑∞n=1(−1)n(2n+1)22S=−∑∞m=0(−1)m(2m+1)2−A+∑∞n=1(−1)n(2n+1)2withA=∑∞n=12(2n+1)(2n−1)2S=−1−AletstateUn=∑nk=12(2k+1)(2k−1)wehaveA=limn→∞UnUn=∑nk=1(12k−1−12k+1)=∑n−1k=012k+1−∑nk=112k+1=1−12n+1SoA=1thenS=−2 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: calculate-0-cos-2pix-x-2-3-2-dx-Next Next post: findf-a-arctan-cosx-x-2-a-2-dx-witha-gt-0- Leave a Reply Cancel replyYour email address will not be published.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.