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


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Question Number 65297 by mathmax by abdo last updated on 28/Jul/19

$l e t U_{n} a s e q u e n c e w i c h v e r i f y U_{n} + U_{n + 1} + U_{n + 2} = n {(- 1)}^{n}$ $f o r a l l i n t e g r n c a l c u l a t e i n t e r m s o f n$ $A_{n} = \sum_{k = 0}^{n} {(- 1)}^{k} U_{k}$ $t h e f i r s t t e r m i s U_{0}$
Commented by mathmax by abdo last updated on 29/Jul/19

$w e h a v e u_{n} + u_{n + 1} + u_{n + 2} = n {(- 1)}^{n} \Rightarrow$ $\sum_{k = 0}^{n} {(- 1)}^{k} (u_{k} + u_{k + 1}) + \sum_{k = 0}^{n} {(- 1)}^{k} u_{k + 2} = \sum_{k = 0}^{n} k {(- 1)}^{k}$ $\sum_{k = 0}^{n} {(- 1)}^{k} (u_{k} + u_{k + 1}) = u_{0} + u_{1} - u_{1} - u_{2} + . . . . {(- 1)}^{n - 1} (u_{n - 1} + u_{n})$ $+ {(- 1)}^{n} (u_{n} + u_{n + 1}) = u_{0} + {(- 1)}^{n} u_{n + 1} \Rightarrow$ $\sum_{k = 0}^{n} {(- 1)}^{k} u_{k + 2} = \sum_{k = 0}^{n} k {(- 1)}^{k} - u_{0} - {(- 1)}^{n} u_{n + 1}$ $\sum_{k = 2}^{n + 2} {(- 1)}^{k - 2} u_{k} = \sum_{k = 0}^{n} k {(- 1)}^{k} - u_{0} - {(- 1)}^{n} u_{n + 1} \Rightarrow$ $\sum_{k = 0}^{n + 2} {(- 1)}^{k} u_{k} - u_{0} + u_{1} = \sum_{k = 0}^{n} k {(- 1)}^{k} - u_{0} - {(- 1)}^{n} u_{n + 1} \Rightarrow$ $\sum_{k = 0}^{n} {(- 1)}^{k} u_{k} + {(- 1)}^{n + 1} u_{n + 1} + {(- 1)}^{n + 2} u_{n + 2} =$ $\sum_{k = 0}^{n} k {(- 1)}^{k} - {(- 1)}^{n} u_{n + 1} - u_{1} \Rightarrow$ $A_{n} = \sum_{k = 0}^{n} k {(- 1)}^{k} - {(- 1)}^{n} u_{n + 1} + {(- 1)}^{n} u_{n + 1} - {(- 1)}^{n} u_{n + 2} - u_{1}$ $= \sum_{k = 0}^{n} k {(- 1)}^{k} - {(- 1)}^{n} u_{n + 2} - u_{1}$ $l e t p (x) = \sum_{k = 0}^{n} x^{k} \Rightarrow p^{^{'}} (x) = \sum_{k = 1}^{n} {k x}^{k - 1} \Rightarrow {x p}^{^{'}} (x) = \sum_{k = 1}^{n} {k x}^{k}$ $b u t p (x) = \frac{x^{n + 1} - 1}{x - 1} (x \neq 1) \Rightarrow p^{^{'}} (x) = \frac{(n + 1) x^{n} (x - 1) - (x^{n + 1} - 1)}{{(x - 1)}^{2}}$ $= \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow \sum_{k = 0}^{n} {k x}^{k} = \frac{{n x}^{n + 2} - (n + 1) x^{n + 1} + x}{{(x - 1)}^{2}}$ $x = - 1 \Rightarrow \sum_{k = 0}^{n} k {(- 1)}^{k} = \frac{n {(- 1)}^{n + 2} - (n + 1) {(- 1)}^{n + 1} - 1}{4}$ $= \frac{n {(- 1)}^{n} + (n + 1) {(- 1)}^{n} - 1}{4} = \frac{(2 n + 1) {(- 1)}^{n} - 1}{4} \Rightarrow$ $A_{n} = \frac{(2 n + 1) {(- 1)}^{n} - 1}{4} - {(- 1)}^{n} u_{n + 2} - u_{1}$
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