U-n-isa-sequence-woch-verify-U-n-U-n-1-n-2-1-n-n-0-1-detdrmine-U-n-interm-of-n-2-find-nsture-of-the-serie-U-n-n-4-3-calculate-k-j-n-U-k-U-j-

Question Number 77754 by abdomathmax last updated on 09/Jan/20

U_{n} i s a s e q u e n c e w o c h v e r i f y

U_{n} + U_{n + 1} = n^{2} {(- 1)}^{n} \forall n ⩾ 0

1) d e t d r m i n e U_{n} i n t e r m o f n

2) f i n d n s t u r e o f t h e s e r i e Σ \frac{U_{n}}{n^{4}}

3) c a l c u l a t e \sum_{k + j = n} U_{k} U_{j}

Commented by mathmax by abdo last updated on 10/Jan/20

w e h a v e U_{n} + U_{n + 1} = n^{2} {(- 1)}^{n} \Rightarrow

\sum_{k = 0}^{n - 1} {(- 1)}^{k} (U_{k} + U_{k + 1}) = \sum_{k = 0}^{n - 1} {(- 1)}^{k} k^{2} \Rightarrow

U_{0} + U_{1} - (U_{1} + U_{2}) + \dots . {(- 1)}^{n - 2} (U_{n - 2} + U_{n - 1}) + {(- 1)}^{n - 1} (U_{n - 1} + U_{n})

\sum_{k = 0}^{n - 1} k^{2} {(- 1)}^{k} \Rightarrow U_{0} + {(- 1)}^{n - 1} U_{n} = \sum_{k = 0}^{n - 1} k^{2} {(- 1)}^{k} \Rightarrow

{(- 1)}^{n - 1} U_{n} = \sum_{k = 0}^{n - 1} k^{2} {(- 1)}^{k} - U_{0} \Rightarrow

U_{n} = {(- 1)}^{n - 1} \sum_{k = 0}^{n - 1} k^{2} {(- 1)}^{k} - {(- 1)}^{n - 1} U_{0}

= {(- 1)}^{n} U_{0} - {(- 1)}^{n} \sum_{k = 0}^{n - 1} k^{2} {(- 1)}^{k}

l e t p (x) = 1 + x + x^{2} + \dots + x^{n - 1} = \sum_{k = 0}^{n - 1} x^{k} \Rightarrow p (x) = \frac{x^{n} - 1}{x - 1} (x \neq 1)

\Rightarrow \sum_{k = 1}^{n - 1} k x^{k - 1} = \frac{{n x}^{n - 1} (x - 1) - (x^{n} - 1) \times 1}{{(x - 1)}^{2}} =

= \frac{{n x}^{n} - {n x}^{n - 1} - x^{n} + 1}{{(x - 1)}^{2}} = \frac{(n - 1) x^{n} - {n x}^{n - 1} + 1}{{(x - 1)}^{2}} \Rightarrow

\sum_{k = 0}^{n - 1} k x^{k} = \frac{x}{{(x - 1)}^{2}} {(n - 1) x^{n} - {n x}^{n - 1} + 1} = λ (x) \Rightarrow

\sum_{k = 1}^{n - 1} k^{2} x^{k - 1} = λ^{^{'}} (x) \Rightarrow \sum_{k = 0}^{n - 1} k^{2} x^{k} = x λ^{^{'}} (x) \Rightarrow

\sum_{k = 0}^{n - 1} k^{2} {(- 1)}^{k} = - λ^{^{'}} (- 1) \Rightarrow

U_{n} = {(- 1)}^{n} U_{o} + {(- 1)}^{n} λ^{^{'}} (- 1) w e h a v e

λ (x) = \frac{(n - 1) x^{n + 1} - {n x}^{n} + x}{{(x - 1)}^{2}} \Rightarrow

λ^{^{'}} (x) = \frac{{(n^{2} - 1) x^{n} - n^{2} x^{n - 1} + 1) {(x - 1)}^{2} - 2 (x - 1) {(n - 1) x^{n + 1} - {n x}^{n} + x}}{{(x - 1)}^{4}}

= \frac{(x - 1) {(n^{2} - 1) x^{n} - n^{2} x^{n - 1} + 1} - 2 {(n - 1) x^{n + 1} - {n x}^{n} + 1}}{{(x - 1)}^{3}} \Rightarrow

λ^{^{'}} (- 1) = \frac{(- 2) {(n^{2} - 1) {(- 1)}^{n} - n^{2} {(- 1)}^{n - 1} + 1} - 2 {(n - 1) {(- 1)}^{n + 1} - n {(- 1)}^{n} + 1}}{- 8}

= \frac{(2 n^{2} - 1) {(- 1)}^{n} + 1 - 2 {(- 2 n + 1) {(- 1)}^{n} + 1}}{4}

= \frac{(2 n^{2} + 4 n - 3) {(- 1)}^{n} - 1}{4} \Rightarrow

U_{n} = {(- 1)}^{n} U_{0} + {(- 1)}^{n} \times \frac{- 1 + (2 n^{2} + 4 n - 3) {(- 1)}^{n}}{4} f o r n ⩾ 1

Commented by mr W last updated on 10/Jan/20

p l e a s e c h e c k s i r :

i f U_{n} = {(- 1)}^{n} U_{0} + {(- 1)}^{n} \times \frac{- 1 + (2 n^{2} + 4 n - 3) {(- 1)}^{n}}{4}

U_{n + 1} = {(- 1)}^{n + 1} U_{0} + {(- 1)}^{n + 1} \times \frac{- 1 + [2 {(n + 1)}^{2} + 4 (n + 1) - 3] {(- 1)}^{n + 1}}{4}

U_{n + 1} = - {(- 1)}^{n} U_{0} + {(- 1)}^{n} \times \frac{1 + [2 n^{2} + 8 n + 3] {(- 1)}^{n}}{4}

\Rightarrow U_{n + 1} + U_{n} = {(- 1)}^{2 n} (n^{2} + 3 n) \neq {(- 1)}^{n} n^{2}

Commented by mathmax by abdo last updated on 11/Jan/20

p e r h a p s m y a n s w e r c o n t a i n a e r r o r o f c a l c u l u s b u t s i r m r w

g i v e o p p o r t o n i t y t o t h e m e t h o d \dots

Answered by mr W last updated on 10/Jan/20

s a y U_{n} = V_{n} + {(- 1)}^{n - 1} [{a n}^{3} + {b n}^{2} + c n]

U_{n + 1} = V_{n + 1} + {(- 1)}^{n} [a {(n + 1)}^{3} + b {(n + 1)}^{2} + c (n + 1)]

U_{n + 1} + U_{n} = V_{n + 1} + V_{n} + {(- 1)}^{n} {a [- n^{3} + {(n + 1)}^{3}] + b [- n^{2} + {(n + 1)}^{2}] + c [- n + (n + 1)]}

U_{n + 1} + U_{n} = V_{n + 1} + V_{n} + {(- 1)}^{n} {a [3 n^{2} + 3 n + 1] + b [2 n + 1] + c}

U_{n + 1} + U_{n} = V_{n + 1} + V_{n} + {(- 1)}^{n} {3 {a n}^{2} + (3 a + 2 b) n + (a + b + c)} = {(- 1)}^{n} n^{2}

3 a = 1 \Rightarrow a = \frac{1}{3}

3 a + 2 b = 0 \Rightarrow b = - \frac{1}{2}

a + b + c = 0 \Rightarrow c = - \frac{1}{3} + \frac{1}{2} = \frac{1}{6}

\Rightarrow V_{n + 1} + V_{n} = 0 \Rightarrow V_{n + 1} = - V_{n}

\Rightarrow V_{n} = {(- 1)}^{n} V_{0}

\Rightarrow U_{n} = V_{n} + {(- 1)}^{n - 1} [\frac{1}{3} n^{3} - \frac{1}{2} n^{2} + \frac{1}{6} n]

\Rightarrow U_{n} = {(- 1)}^{n} [V_{0} - \frac{n (n - 1) (2 n - 1)}{6}]

a s s u m e U_{0} = 1, \Rightarrow V_{0} = U_{0} = 1

\Rightarrow U_{n} = {(- 1)}^{n} [1 - \frac{n (n - 1) (2 n - 1)}{6}]

c h e c k :

U_{n + 1} = {(- 1)}^{n + 1} [1 - \frac{(n + 1) n (2 n + 1)}{6}]

U_{n + 1} = {(- 1)}^{n} [- 1 + \frac{(n + 1) n (2 n + 1)}{6}]

U_{n + 1} + U_{n} = {(- 1)}^{n} [- \frac{n (n - 1) (2 n - 1)}{6} + \frac{(n + 1) n (2 n + 1)}{6}]

U_{n + 1} + U_{n} = {(- 1)}^{n} n^{2} \Rightarrow o k

Commented by mr W last updated on 10/Jan/20

i s m y m e t h o d o k ?

t h i s i s m y f i r s t t r y t o s o l v e s u c h a k i n d

o f q u e s t i o n s .

Facebook Tweet Pin