find-k-0-n-C-n-k-cos-kpi-n- Tinku Tara June 4, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 90749 by abdomathmax last updated on 25/Apr/20 find∑k=0nCnkcos(kπn) Commented by mathmax by abdo last updated on 26/Apr/20 An=∑k=0nCnkcos(kπn)⇒An=Re(∑k=0nCnkeikπn)wehave∑k=0nCnkeikπn=∑k=0nCnk(eiπn)k=(1+eiπn)n=(1+cos(πn)+isin(πn))n=(2cos2(π2n)+2isin(π2n)cos(π2n)n=(2cos(π2n)eiπ2n)n=2ncosn(π2n)eiπ2=2ncosn(π2n)i⇒An=0 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: cosec-4x-2-dx-Next Next post: calculste-lim-n-1-n-2-k-1-n-karctan-k-n- Leave a Reply Cancel replyYour email address will not be published.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.