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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-




Question Number 65297 by mathmax by abdo last updated on 28/Jul/19
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 =Σ_(k=0) ^n  (−1)^k  U_k   the first term is U_0
letUnasequencewichverifyUn+Un+1+Un+2=n(1)nforallintegrncalculateintermsofnAn=k=0n(1)kUkthefirsttermisU0
Commented by mathmax by abdo last updated on 29/Jul/19
we have u_n  +u_(n+1)  +u_(n+2) =n(−1)^n  ⇒  Σ_(k=0) ^n (−1)^k  (u_k  +u_(k+1) )+Σ_(k=0) ^n (−1)^k  u_(k+2) =Σ_(k=0) ^n  k(−1)^k   Σ_(k=0) ^n (−1)^k (u_k  +u_(k+1) ) =u_0 +u_1 −u_1 −u_2  +....(−1)^(n−1) (u_(n−1)  +u_n )  +(−1)^n (u_n  +u_(n+1) ) =u_0  +(−1)^n u_(n+1) ⇒  Σ_(k=0) ^n (−1)^k u_(k+2) =Σ_(k=0) ^n k(−1)^k −u_0 −(−1)^n u_(n+1)   Σ_(k=2) ^(n+2) (−1)^(k−2)  u_k =Σ_(k=0) ^n k(−1)^k −u_0 −(−1)^n u_(n+1) ⇒  Σ_(k=0) ^(n+2) (−1)^k u_k −u_0 +u_1 =Σ_(k=0) ^n k(−1)^k −u_0 −(−1)^n u_(n+1)  ⇒  Σ_(k=0) ^n  (−1)^k  u_k   +(−1)^(n+1) u_(n+1) +(−1)^(n+2) u_(n+2) =  Σ_(k=0) ^n k(−1)^k −(−1)^n u_(n+1) −u_1 ⇒  A_n =Σ_(k=0) ^n k(−1)^k −(−1)^n u_(n+1) +(−1)^n u_(n+1) −(−1)^n u_(n+2)  −u_1   =Σ_(k=0) ^n k(−1)^k   −(−1)^n u_(n+2) −u_1   let p(x) =Σ_(k=0) ^n  x^k  ⇒p^′ (x) =Σ_(k=1) ^n kx^(k−1)  ⇒xp^′ (x) =Σ_(k=1) ^n kx^k   but p(x) =((x^(n+1) −1)/(x−1)) (x≠1) ⇒p^′ (x) =(((n+1)x^n (x−1)−(x^(n+1) −1))/((x−1)^2 ))  =((nx^(n+1) −(n+1)x^n  +1)/((x−1)^2 )) ⇒Σ_(k=0) ^n kx^k  =((nx^(n+2) −(n+1)x^(n+1)  +x)/((x−1)^2 ))  x=−1 ⇒Σ_(k=0) ^n  k(−1)^k  =((n(−1)^(n+2) −(n+1)(−1)^(n+1) −1)/4)  =((n(−1)^n +(n+1)(−1)^n −1)/4) =(((2n+1)(−1)^n −1)/4) ⇒  A_n =(((2n+1)(−1)^n −1)/4) −(−1)^n u_(n+2)  −u_1
wehaveun+un+1+un+2=n(1)nk=0n(1)k(uk+uk+1)+k=0n(1)kuk+2=k=0nk(1)kk=0n(1)k(uk+uk+1)=u0+u1u1u2+.(1)n1(un1+un)+(1)n(un+un+1)=u0+(1)nun+1k=0n(1)kuk+2=k=0nk(1)ku0(1)nun+1k=2n+2(1)k2uk=k=0nk(1)ku0(1)nun+1k=0n+2(1)kuku0+u1=k=0nk(1)ku0(1)nun+1k=0n(1)kuk+(1)n+1un+1+(1)n+2un+2=k=0nk(1)k(1)nun+1u1An=k=0nk(1)k(1)nun+1+(1)nun+1(1)nun+2u1=k=0nk(1)k(1)nun+2u1letp(x)=k=0nxkp(x)=k=1nkxk1xp(x)=k=1nkxkbutp(x)=xn+11x1(x1)p(x)=(n+1)xn(x1)(xn+11)(x1)2=nxn+1(n+1)xn+1(x1)2k=0nkxk=nxn+2(n+1)xn+1+x(x1)2x=1k=0nk(1)k=n(1)n+2(n+1)(1)n+114=n(1)n+(n+1)(1)n14=(2n+1)(1)n14An=(2n+1)(1)n14(1)nun+2u1

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