let-U-n-a-sequence-wich-verify-U-n-U-n-1-U-n-2-n-1-n-for-all-integr-n-calculate-interms-of-n-A-n-k-0-n-1-k-U-k-the-first-term-is-U-0- Tinku Tara June 4, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 65297 by mathmax by abdo last updated on 28/Jul/19 letUnasequencewichverifyUn+Un+1+Un+2=n(−1)nforallintegrncalculateintermsofnAn=∑k=0n(−1)kUkthefirsttermisU0 Commented by mathmax by abdo last updated on 29/Jul/19 wehaveun+un+1+un+2=n(−1)n⇒∑k=0n(−1)k(uk+uk+1)+∑k=0n(−1)kuk+2=∑k=0nk(−1)k∑k=0n(−1)k(uk+uk+1)=u0+u1−u1−u2+….(−1)n−1(un−1+un)+(−1)n(un+un+1)=u0+(−1)nun+1⇒∑k=0n(−1)kuk+2=∑k=0nk(−1)k−u0−(−1)nun+1∑k=2n+2(−1)k−2uk=∑k=0nk(−1)k−u0−(−1)nun+1⇒∑k=0n+2(−1)kuk−u0+u1=∑k=0nk(−1)k−u0−(−1)nun+1⇒∑k=0n(−1)kuk+(−1)n+1un+1+(−1)n+2un+2=∑k=0nk(−1)k−(−1)nun+1−u1⇒An=∑k=0nk(−1)k−(−1)nun+1+(−1)nun+1−(−1)nun+2−u1=∑k=0nk(−1)k−(−1)nun+2−u1letp(x)=∑k=0nxk⇒p′(x)=∑k=1nkxk−1⇒xp′(x)=∑k=1nkxkbutp(x)=xn+1−1x−1(x≠1)⇒p′(x)=(n+1)xn(x−1)−(xn+1−1)(x−1)2=nxn+1−(n+1)xn+1(x−1)2⇒∑k=0nkxk=nxn+2−(n+1)xn+1+x(x−1)2x=−1⇒∑k=0nk(−1)k=n(−1)n+2−(n+1)(−1)n+1−14=n(−1)n+(n+1)(−1)n−14=(2n+1)(−1)n−14⇒An=(2n+1)(−1)n−14−(−1)nun+2−u1 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: 1-calculate-dx-x-a-with-a-C-2-find-the-values-of-0-dx-x-4-1-and-0-dx-x-6-1-by-using-the-decomposition-inside-C-x-Next Next post: calculate-0-1-ln-1-x-4-x-2-dx- Leave a Reply Cancel replyYour email address will not be published.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.