let u_n =Σ_(k=0) ^n (3k+1)(−1)^k 1) calculate interms of n S_n =u_0 +u_1 +u_2 +....+u_n 2) calculate u_0 +u_1 +u_2 +....+u


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Question Number 40370 by math khazana by abdo last updated on 20/Jul/18

$l e t u_{n} = \sum_{k = 0}^{n} (3 k + 1) {(- 1)}^{k}$ $1) c a l c u l a t e i n t e r m s o f n$ $S_{n} = u_{0} + u_{1} + u_{2} + . . . . + u_{n}$ $2) c a l c u l a t e u_{0} + u_{1} + u_{2} + . . . . + u_{57}$
Commented by prof Abdo imad last updated on 20/Jul/18
$1)we have u_n =3Σ_(k=0) ^n k(−1)^(k ) +Σ_(k=0) ^n (−1)^k but Σ_(k=0) ^n (−1)^k =((1−(−1)^(n+1) )/2) =((1+(−1)^n )/2) let p(x)=Σ_(k=0) ^n x^k with x≠1 we have p^′ (x)=Σ_(k=1) ^n k x^(k−1) ⇒x p^′ (x)=Σ_(k=1) ^n k x^k ⇒ Σ_(k=1) ^n k(−1)^k =−p^′ (−1) but p(x)=((x^(n+1) −1)/(x−1)) ⇒ p^′ (x) =((nx^(n+1) −(n+1)x^n +1)/((x−1)^2 )) ⇒ p^′ (−1) =((n(−1)^(n+1) −(n+1)(−1)^n +1)/4) ⇒ u_n =−(3/4){ 1−(2n+1)(−1)^n } +((1+(−1)^n )/2) u_n =(3/4){(2n+1)(−1)^n −1} +((1+(−1)^n )/2) we have S_n =Σ_(k=0) ^n u_k =(3/4) Σ_(k=0) ^n (2k+1)(−1)^k −(3/4)Σ_(k=0) ^n (1) +(1/2)Σ_(k=0) ^n (1) +(1/2)Σ_(k=0) ^n (−1)^k =(3/2)Σ_(k=0) ^n k(−1)^k +(3/4)Σ_(k=0) ^n (−1)^k −(3/4)(n+1) +((n+1)/2) +(1/2) ((1+(−1)^n )/2) =(3/2){((1−(2n+1)(−1)^n )/4)} +(3/4) ((1+(−1)^n )/2) −(1/4)(n+1) +(1/4){1+(−1)^n } S_n =(3/8){1−(2n+1)(−1)^n } +(5/8){1+(−1)^n } −((n+1)/4) .$
$1) w e h a v e u_{n} = 3 \sum_{k = 0}^{n} k {(- 1)}^{k} + \sum_{k = 0}^{n} {(- 1)}^{k} b u t$ $\sum_{k = 0}^{n} {(- 1)}^{k} = \frac{1 - {(- 1)}^{n + 1}}{2} = \frac{1 + {(- 1)}^{n}}{2}$ $l e t p (x) = \sum_{k = 0}^{n} x^{k} w i t h x \neq 1 w e h a v e$ $p^{^{'}} (x) = \sum_{k = 1}^{n} k x^{k - 1} \Rightarrow x p^{^{'}} (x) = \sum_{k = 1}^{n} k x^{k} \Rightarrow$ $\sum_{k = 1}^{n} k {(- 1)}^{k} = - p^{^{'}} (- 1) b u t p (x) = \frac{x^{n + 1} - 1}{x - 1} \Rightarrow$ $p^{^{'}} (x) = \frac{{n x}^{n + 1} - (n + 1) x^{n} + 1}{{(x - 1)}^{2}} \Rightarrow$ $p^{^{'}} (- 1) = \frac{n {(- 1)}^{n + 1} - (n + 1) {(- 1)}^{n} + 1}{4} \Rightarrow$ $u_{n} = - \frac{3}{4} {1 - (2 n + 1) {(- 1)}^{n}} + \frac{1 + {(- 1)}^{n}}{2}$ $u_{n} = \frac{3}{4} {(2 n + 1) {(- 1)}^{n} - 1} + \frac{1 + {(- 1)}^{n}}{2} w e h a v e$ $S_{n} = \sum_{k = 0}^{n} u_{k}$ $= \frac{3}{4} \sum_{k = 0}^{n} (2 k + 1) {(- 1)}^{k} - \frac{3}{4} \sum_{k = 0}^{n} (1)$ $+ \frac{1}{2} \sum_{k = 0}^{n} (1) + \frac{1}{2} \sum_{k = 0}^{n} {(- 1)}^{k}$ $= \frac{3}{2} \sum_{k = 0}^{n} k {(- 1)}^{k} + \frac{3}{4} \sum_{k = 0}^{n} {(- 1)}^{k} - \frac{3}{4} (n + 1)$ $+ \frac{n + 1}{2} + \frac{1}{2} \frac{1 + {(- 1)}^{n}}{2}$ $= \frac{3}{2} {\frac{1 - (2 n + 1) {(- 1)}^{n}}{4}} + \frac{3}{4} \frac{1 + {(- 1)}^{n}}{2}$ $- \frac{1}{4} (n + 1) + \frac{1}{4} {1 + {(- 1)}^{n}}$ $S_{n} = \frac{3}{8} {1 - (2 n + 1) {(- 1)}^{n}} + \frac{5}{8} {1 + {(- 1)}^{n}}$ $- \frac{n + 1}{4} .$
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